I was thinking about how would capillary action change in a tube (classic example) and in a tube fitted inside another tube (considering water as the liquid involved).
Height of liquid column:
where:
$\gamma$ = liquid-air surface tension
$\theta$ = contact angle
$\rho$ = density of liquid
$g$ = gravity acceleration
$r$ = radius
I tried my best to draw the examples I'm interested in order to help my explanation.
I didn't consider the capillarity inside the smaller tube in both example #2 and #3 because I'd like to assume that $a/2$ in example #1 is close to $c$ in example #2 and #3 (drawings not to scale).
Since from what I understand the column height is given, among other things (most of which can't be changed, like liquid-air surface tension, contact angle, density of liquid and gravity acceleration), by the tube radius, I'd like to know if $c$ in example #2 can be considered as $a/2$ in example #1 to calculate column height using above formula.
Also I'd like to know how having beads of slightly smaller diameter than $c$ between the two tubes (example #3) would affect the column height.
If said beads were less dense than water, could they still improve column height or would they just form a floating mat on top of 1 unit thickness?
What'd be the column height of example #2 and #3 assuming $c$ as 1mm?
I'm quite sure that given the same reached height $h$ in example #2 and #3, $c$ of #2 has to be smaller than $c$ in #3.
Best Answer
If $c\ll R$ the radius of your outer tube, the total curvature is approximately $\cos \theta/c$, so you will get $$h = \frac{\gamma}{\rho g} \frac{\cos \theta}{c}.$$
Beads will usually lower the apparent surface tension, so you'll get a lower column, although the amount of that depends on their wetting properties and of their arrangement (packing), see e.g. http://www.sciencedirect.com/science/article/pii/S0021979785711502