Timeline for How to find out the maximum radius of a hole that can keep water stay in a container by water viscosity?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2018 at 18:52 | comment | added | John Rennie | @wesanyer See this Wikipedia article | |
| Jun 15, 2018 at 18:04 | comment | added | wesanyer | Is this $\gamma$ a material property? Or is it calculated based on certain material properties? | |
| Jun 15, 2018 at 13:17 | comment | added | John Rennie | @wesanyer: $\gamma$ is the surface tension at the air/water interface not the specific weight | |
| Jun 15, 2018 at 12:31 | comment | added | wesanyer | According to wikipedia, the specific weight $\gamma$ should be equal to $\rho g$. Therefore, simplifying your equation: $r = \frac{2\gamma}{\rho g h} = \frac{2 \rho g}{\rho g h} = \frac{2}{h}$. But then your example doesn't work: I get 2 mm as the radius for a cone height of 10 cm, not 0.1mm. Am I looking at this wrong? | |
| Jan 23, 2018 at 11:18 | comment | added | Gopinath | What about the wetting angle? Why is it not part of the equation? | |
| Jul 3, 2016 at 17:46 | comment | added | Mrigank | Sorry.... just got it, to fit in the hole, radius of curvature in laplace equation must be more than or equal to radius of hole.. | |
| Jul 3, 2016 at 17:23 | comment | added | John Rennie | @ELiT: I'm not sure what you're asking about the radius of curvature. The highest pressure within the forming droplet is when the droplet is a hemisphere with radius equal to the hole radius. My calculation is to equate this maximum pressure with the pressure at the bottom of the cone. | |
| Jul 3, 2016 at 17:21 | comment | added | Mrigank | about radius of curvature? if we are ignoring weight, then radius of curvature can literally be anything so, it can tend zero and height may be infinite... | |
| Jul 3, 2016 at 17:05 | comment | added | John Rennie | @ELiT: ignoring the weight of the drop is an approximation, but in most circumstances it's a very good approximation. | |
| Jul 3, 2016 at 17:01 | comment | added | Mrigank | I would like to ask, how did you determine radius of curvature of drop?, I mean how to prove mathematically that it's the maxima/minima... and weight of drop was not included? | |
| Dec 18, 2012 at 2:32 | vote | accept | Marco | ||
| Dec 17, 2012 at 10:54 | history | answered | John Rennie | CC BY-SA 3.0 |