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I'm looking to calculate the diameter of a liquid after being poured on a flat surface, in terms of: Time - t Viscosity - η Density - ρ Volume poured - V

Basically, assume you were to carefully pour a liquid (ex. water, milk, oil) onto a flat surface (ex. a glass sheet) and measure the diameter of the puddle over time. If we know the viscosity, density, and volume poured, how could we calculate (or estimate) the diameter over time?

Assume it's a Newtonian fluid. This would be for a larger scale as well, like pouring a bucket of water on a floor, not a drop of water, so we can presumably ignore surface tension?

Thank you!

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    $\begingroup$ I'm no physicist but I don't think you can ignore surface tension otherwise what stops it from just forming a constant depth layer (which would be zero thickness if there are no upward edges around thew surface to contain it)? $\endgroup$
    – DKNguyen
    Commented Sep 4, 2020 at 18:17
  • $\begingroup$ Fair enough, good catch. Could we reasonably assume surface tension won't affect the spread rate while the depth is at least an order of magnitude higher than the depth once the flow stops? For example, while the liquid depth is 5cm? $\endgroup$
    – Bongo
    Commented Sep 5, 2020 at 11:11
  • $\begingroup$ I would imagine this depends on how dense the liquid relative to its surface tension (which I imagine is closely tied to its viscosity) is since a very heavy liquid requires a lot less volume in for its viscous forces to dominate over the inertial forces. $\endgroup$
    – DKNguyen
    Commented Sep 5, 2020 at 15:15
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    $\begingroup$ See the Spreading Dynamics section of this article: en.wikipedia.org/wiki/Wetting?wprov=sfla1 $\endgroup$
    – electronpusher
    Commented Sep 6, 2020 at 16:52

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A quick and dirty estimate of the radius $r$ is $$ r=\sqrt{\frac{V}{\pi h}} $$ where $$ h=\sqrt{\frac{2\sigma(1-\cos\theta)}{\rho g}} $$

besides the already given symbols, $\sigma$ is the surface tension, $g$ acceleration due to gravity, $\theta$ the contact angle b/w the liquid and surface and $h$ the height of the puddle. $V$ would be a function of time in terms of flow rate.

The actual equivalent radius must be a statistical average for a given V at equilibrium.

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  • $\begingroup$ Thanks for the answers! I realize I should elaborate more, sorry for that. I'm assuming the "puddle" is very large, and still flowing. I'm less interested in how large it will be when it stops, and more interested how it will grow with time. I'm also assuming we pour the liquid at once and then watch how it spreads. For a visual, imagine dumping 100L of honey on an large glass plate. Then watching it spread. How could we approximate the diameter let's from, let's say, the 1min mark to the 5min mark. $\endgroup$
    – Bongo
    Commented Sep 5, 2020 at 10:59

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