# Capillary length and Bond number confusion

The Bond number represents the ratio of gravity forces to surface tension forces, and is defined as

$$Bo = \frac{\rho g L^2}{\sigma}$$

where $$\rho$$ is the fluid density, $$g$$ is gravitational acceleration, $$L$$ is a characteristic length of the surface, and $$\sigma$$ is the surface tension.

When gravity and surface tension forces balance ($$Bo \sim 1$$), we can solve for the capillary rise on a surface:

$$L \sim \sqrt{\frac{\sigma}{\rho g}}$$

However, if $$Bo >> 1$$ (gravity dominates and/or weak surface tension), this implies $$L >> \sqrt{\frac{\sigma}{\rho g}}$$.

Why do we get a larger capillary length for larger Bond number? Shouldn't the capillary length be small if surface tension effects are negligible compared to gravity?

I think you may be confusing the characteristic (capillary) length with the height of the liquid at the surface (the capillary rise). If $$L$$ is the characteristic length of the surface of the liquid, then it would more properly be the length at the edge of the liquid, or the length at the edge where capillary forces occur. Note that the capillary force acting at the edge is $$F=\sigma L$$, where $$L$$ is the length of the edge (considering there is air only on the upper side of the liquid surface).

Then things would fit in. For a capillary tube of radius $$R$$, perfectly wetted by the liquid, the characteristic length is its circumference $$L=2 \pi R$$ and the capillary force would act on $$L$$ and be oriented upwards:

$$F=\sigma L = \sigma \cdot 2 \pi R$$.

The capillary pressure is force over transversal surface of liquid inside the tube $$S=\pi R^2$$:

$$p=\frac{F}{S}=\frac{2 \sigma}{R}$$

Make this equal to the gravitational pressure of the liquid that rises inside the tube at height $$h$$:

$$\rho g h = \frac{2 \sigma}{R}$$,

$$h = \frac{2 \sigma}{\rho g R} = \frac{4 \pi \sigma}{\rho g L}$$.

Now if you take $$L$$ out of the Bond number an insert it in $$h$$:

$$L= \sqrt{\frac{Bo \cdot \sigma}{\rho g}}$$,

$$h = 4 \pi \sqrt{\frac{\sigma}{\rho g \cdot Bo}}$$.

If gravity dominates then $$Bo$$ is higher and the capillary height $$h$$ is lower. If gravity does not dominate but you liked it to dominate then you would have to increase $$L$$ or the radius of the tube $$R$$, which is correct.

• I thought that capillary rise was the capillary length in the Bond number, but I guess this isn't true? – Drew Dec 31 '18 at 18:48
• Frankly I don't think the Bond number contains a very specific characteristic length. The way it looks, $Bo=G/Fsurface=\rho V g / (\sigma L) = \rho g L^3 / (\sigma L) = \rho g L^2 / \sigma$. This means the body of liquid is taken to be a cube and the surface force acts on one side of it. This seems to be a rough way of looking at surface forces, not a very exact one. That is good though, because this approach roughly works in many cases. But then you'd have to choose L in a proper way. I'd say it's the contour on which surface forces act. In turn the rise is their effect. – Ferred Dec 31 '18 at 21:23

Quote from Here:

"The Capillary number describes the ratio of viscous to capillary forces, while the Bond number indicates the ratio of gravitational to capillary forces". Additional links

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