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I When to use which dimensionless number

  1. Dec 1, 2017 #1

    joshmccraney

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    Hi PF!

    I've been reading about low gravity capillary driven flows, and no authors use Reynolds number when measuring importance of inertia in capillary driven flows. Instead most use the Ohnesorge number. Can someone explain why this is?

    Thanks!
     
  2. jcsd
  3. Dec 1, 2017 #2

    BvU

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    Did you google ?
     
  4. Dec 1, 2017 #3

    joshmccraney

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    I did. The Ohnesorge number is evidently like a Reynolds number but with a capillary length scale. What would cause someone to think of this?
     
  5. Dec 1, 2017 #4

    BvU

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    Significance of surface tension in capillary flow
     
  6. Dec 1, 2017 #5

    joshmccraney

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    But there doesn't seem to be anything intrinsically wrong with Reynolds number. Sure, it doesn't measure surface tension, but why isn't it really ever used for low Bond number capillary flows?
     
  7. Dec 1, 2017 #6

    BvU

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    @Chestermiller : you know of someone in PF with specific expertise here ? Unfortunately for Josh, mine is just hearsay :sorry: .
     
  8. Dec 1, 2017 #7
    Very often in systems like this, we start out with the differential equations and use a systematic approach developed by Hellums and Churchill to reduce the equations to dimensionless form. The key dimensionless groups automatically emerge from this methodology (with some small leeway on selection of the groups). As first year graduate students at Michigan, we were taught this simple methodology by Stu Churchill back in 1963 (at that time, he was chairman of the ChE Dept.). It has served me well over the years.
     
  9. Dec 1, 2017 #8

    BvU

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    Thanks, Chet !

    Also consulted with my roommate (from my point of view an expert...) who knows all these numbers, but he didn't come much further than: we use it in droplet formation to find correlations.

    My own googling didn't get me far, but I did like the overview lemma.

    As a physicist, when I 'm in a bad mood, I am inclined to think that these chemical engineers make up these dimensionless numbers as much as they can for two reasons: 1. as a claim to fame and immortality and 2. to make parity plots (preferably log-log) where at least three of these numbers appear with fractional exponents. :wink: But only when I'm in a bad mood, of course. And my roommate is a mathematician.
     
  10. Dec 1, 2017 #9

    joshmccraney

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    hahahahaha this is a funny point! :oldlaugh: Ok then, maybe it's simply from scaling the equations! Thanks for your guys' help!
     
  11. Dec 3, 2017 #10

    boneh3ad

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    Well, the Ohnesorge number is related to the Weber number and the Reynolds number, so under certain conditions (##We\approx 1##) it is effectively equivalent to Reynolds number.

    The greater point here is that the choice of dimensionless group depends on the phenomena being studied at the time. The point made by @Chestermiller is important in that these dimensionless groups fall out of the governing equations (most importantly the Navier-Stokes equations) when you make them dimensionless based on scales that are important to the problem you are studying. However, what those scales are is up to you to define and will change from one class of problems to the next. Alternatively, one can carry out a dimensional analysis of the problem (e.g. by the Buckingham Pi Theorem) and those groups will fall out based on the various combinations of important parameters to the problem.

    The bottom line is that different physics are dominant for different problems, and your choice of dimensionless groupings must reflect that.
     
  12. Dec 3, 2017 #11
    Josh,

    Why don't you provide the dimensional equations for your specific problem, and I'll illustrate the procedure.

    Chet
     
  13. Dec 4, 2017 at 2:01 AM #12

    joshmccraney

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    The problem is fluid flowing down a wedge, though I think I understand the scaling, if that's what you're referring to? If ##z## is the direction of flow, ##y## is the coordinate going into the wedge, and ##x## goes from the corner upwards, ##\alpha## the half-corner angle of the wedge, and ##L## a characteristic length (in ##z## direction) and ##H## characteristic height (in ##x## direction) then scales look like
    $$z \sim L\\
    x \sim H\\
    y \sim H\tan\alpha.$$
    In this problem the Bond number is very small, and gravity can be neglected. Then Young-Laplace governs pressure (we assume negligible surface stress, since air is above the fluid) and the transverse curvature is assumed much greater than axial, reducing pressure equation to:
    $$\Delta P = \sigma\frac{1}{R}\implies\\
    P \sim \frac{\sigma}{fH}$$
    where ##f = (\cos\theta/\sin\alpha-1)^{-1}## is a geometric function such that ##fH=R## where ##R## is the radius of curvature in the ##x-y## plane and ##\theta## is static contact angle of fluid. Then scaling the ##z## component of Navier Stokes yields to order ##\epsilon \equiv H/L##:
    $$P_z = \sin^2\alpha\partial^2_xw+\cos^2\alpha\partial^2_yw.$$
    It makes perfect sense to me that the Ohnesorge number emerges with the inertial terms (not included here for simplicity). However, to get a Reynolds number you'd have the Weber number like boneh3ad said. It's just interesting to me that this fluids problem doesn't really require a Reynolds number. What do you both think?
     
  14. Dec 4, 2017 at 7:14 AM #13
    I think I'd still like to see the formulation of the differential equations and boundary conditions.
     
  15. Dec 4, 2017 at 11:55 AM #14

    joshmccraney

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    Before I continue, I forgot a very important scale: ##w##. If we assume pressure balances viscosity, then dimensional constant density NS yields in ##z## component
    $$\partial_zP = \mu \nabla^2 w\implies\\
    \frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2}+\frac{1}{H^2\tan^2\alpha}+\frac{1}{L^2}\right)w\implies\\
    \frac{\sigma}{H f L} \sim \mu\left(\frac{\tan^2\alpha + 1}{H^2\tan^2\alpha}\right)w\implies\implies\\
    \frac{\sigma}{H f L} \sim \mu\left(\frac{\sec^2\alpha}{H^2\tan^2\alpha}\right)w\implies\\
    \frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2\sin^2\alpha}\right)w\implies\\
    w \sim \frac{\sigma H\sin^2\alpha}{ f L \mu}\implies\\
    w \sim \frac{\epsilon \sigma \sin^2\alpha}{ f \mu}.
    $$
    Notice I neglected ##1/L^2## term since ##\epsilon = H^2/L^2\ll 1## (slender column). Also, I would guess time scales as ##t \sim L/w##, though I'm not sure why I'm inclined to use ##w## rather than ##u,v##. Most of the flow is ##z## directed; is that good enough reason? Anyways, the formulation of the differential equations, assuming we start with dimensional NS and constant density, would be (for ##z## component, though I could do the other two if you'd like as well)

    $$
    \rho\frac{D w}{Dt} = \partial_zP+\mu \nabla^2w\implies\\
    \frac{\rho W}{L/W}\frac{D w}{Dt} = \frac{\sigma}{HfL}\partial_zP+\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w\implies\\
    \frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\frac{HfL}{\sigma}\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w
    $$

    but we already balanced pressure and viscosity, so we know both terms on the RHS are now ##O(1)##, thus we have

    $$\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
    \left(\frac{\epsilon \sigma \sin^2\alpha}{ f \mu}\right)^2\frac{\rho }{1}\frac{Hf}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
    \left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{\sigma \rho H}{\mu^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
    \left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{1}{Oh^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\$$
    and ##Oh \equiv \mu/\sqrt{\rho \sigma H}##. How does this look?
     
  16. Dec 4, 2017 at 12:01 PM #15

    joshmccraney

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    And boundary conditions....oh gosh. So zero slip on the wedge, zero stress along the free surface. Hmmmm physically aren't those it?
     
  17. Dec 5, 2017 at 6:58 AM #16
    I don't understand your description of the physical problem. Can you please provide a diagram?
     
  18. Dec 5, 2017 at 12:39 PM #17

    Andy Resnick

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    But what is driving the flow? An (external) applied pressure gradient? Marangoni effect? Something else?

    Edit: hang on, I re-read your OP. Capillary driven flow. Got it.

    Edit #2: Here's a reference you may find useful, especially sections 2.2 and 2.3: https://ac.els-cdn.com/S03019322060...t=1512499461_7ce3a4db15d949752b591ed73f002381
     
  19. Dec 5, 2017 at 12:45 PM #18

    joshmccraney

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    Here is a schematic of the flow.
     

    Attached Files:

  20. Dec 5, 2017 at 12:46 PM #19
    Hi Andy.

    If you understand the geometry, maybe you can provide a diagram before Josh does.
     
  21. Dec 5, 2017 at 12:51 PM #20

    Andy Resnick

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    This is a 'standard' problem- but not a simple one. There's at least one book and 1 NASA flight experiment that is devoted to this specific problem. Can't immediately find a PDF breaking down the governing equations, but I'm sure the OP can find a suitable reference.
     
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