# Capillary in a slanted tube with a bend

So I've read a question on here about Capillary action in a slanted tube, but my question goes a little further. Firstly though, could someone explain to me what is meant by the only answer to that question. The answer states that the lift force isn't dependant on the weight of the fluid whereas the capillary lift hight equation is a combination of the forces from gravity and capillary action. When I have tried to balance these equation for both a upright tube and a slanted tube I get different lift heights and masses of fluid lifted. I'll explain some more at the bottom of the question.

If we have a slanted tube that slants 15 degree to the right for 10mm and then 15 degree to the left (from normal - not going back to upright)for 10mm would this tube have the same lift height as a tube that slanted 15 degrees for 20mm? (Assuming the tube radius is constant in and equal). Essentially is there any issue with the fluid going around the bend?

As promised more explanation of what I've done thus far.

A tube filled to a given height has the same volume irrespective of the angle the tub is at (see Volume of a slanted tube). So with this in mind and knowing that the side length of the cylinder up to the water level is longer than the height of the water (if the cylinder is slanted), we can comfortably say that to lift the same mass of water the height will be the same, but the length up the pipe will be lower.

However the lifting force from capillary action is based on the circumference of the pipe (2*pi*r), but this changes dependant on angle since a slanted pipe will have a cross section that is a ellipse. So I expect the lifting force to be greater, allowing for a greater amount of fluid to be lifted, thereby increasing the height that the fluid is lifted to.

• OTOH, the volume of fluid per vertical unit increases w/ same proportion as your cross-sectional ellipse of contact area. – Carl Witthoft Apr 29 '14 at 12:44

The curved surface of the liquid in the capillary gives rise to a pressure differential. I believe the shape of the tube below that is irrelevant. The height of the liquid (measured vertically) will be determined by this $\Delta P = \rho \cdot g \cdot \Delta h$