# Under what conditions does Jurin's law work? (and what are its physical basis?)

I searched answer in wikipedia (and other sites) but I found disjointed and contradictory informations. In Jurin's law article they say that the law works if the radius of the tube is smaller than capillary height (by the way, very different statement respect to "much smaller") but in "capillary action" article they use the law to say that, for a 2 m tube radius, water rise is 7 $\mu$m (really this calculus has sense?). Jurin law works for big tube too? If not how can we define the liming case in which it works? I proved (see theorem below) that 3 hypothesis are sufficient to give Jurin law. But they are also necessary? And how could I justify them?

Theorem

If this 3 hypotesis works

i) contact angle $\theta$ is "given", it is a constant depending on liquid and wall material but not dependent by the radius of the tube

ii) for small tubes the meniscus is a part of sphere

iii) for some tube radius $R$, for which $R^2 \ll \frac{2 \sigma \cos^2 \theta}{\rho g}$ ($\sigma$, $\rho$ and $g$ are surface tension, liquid density and gravity acceleration), we have also $h_0 \gg \frac{R}{\cos \theta}$ ($h_0$ height of the bottom of the meniscus)

then, for every smaller $R$, Jurin law works $$h_0 = \frac{2 \sigma \cos \theta}{\rho g R}$$ I don't have justification of this hypothesis, if not the fact that they give the experimentally checked Jurin law (this is a weak justification, I didn't prove the reverse implication).

Proof

I'll suppose that a circular tube is immersed in a big vessel and I'll call "sea" the external liquid (its level far away is independent by all). To proof Jurin law we will write $h(r)$ using the hypothesis i and ii, and we will show that if works hypothesis iii too, the differential equation that rules the surface of the meniscus gives the law for enough small tubes. I showed in the question "What is the meniscus shape?" that this differential equation is $$r h = \ell^2 \left( \frac{ r h''}{(1+(h')^2)^{\frac{3}{2}}} + \frac{h'}{\sqrt{1+(h')^2}} \right)$$ where $\ell = \sqrt{\frac{\sigma}{\rho g}}$.

To find $h(r)$ consider this sketch (where it is in red) of sectioned tube

where $\lambda$ is the radius of curvature of the shape, $\theta$ is the conctact angle, $h_0$ is the bottom of the meniscus surface, the zero $h$ corrisponding to the $r$ axis is the external sea level, $r$ is the distance from the center of the tube. The center of curvature is in $(0,h_0+\lambda)$ while the generic position of the shape of the meniscus is in $(r,h(r))$, so the distance squared (between this two point) is $r^2+(h(r)-h_0-\lambda)^2$. Equaling to $\lambda^2$ we can find $h(r)$, excluding the spurious solution with negative derivative, we get the one physically interesting: $$h(r) = \lambda + h_0 - \sqrt{\lambda^2-r^2}$$ whose first and second derivative can be simplified in this way $$\frac{r}{\sqrt{\lambda^2-r^2}} \qquad \qquad \frac{\lambda^2}{\left( \lambda^2 - r^2 \right)^{\frac{3}{2}}}$$ Substituting into differential equation we finally find $$1 - \sqrt{1-\left( \frac{r}{\lambda} \right)^2} + \frac{h_0}{\lambda} = \frac{2 \ell^2}{\lambda^2}$$ Watching the figure we can write $\lambda$ as function of $\theta$ and $R$ (radius of the tube, i.e. distance between the $h$ axis and the wall): $\lambda=\frac{R}{\cos \theta}$. Substituting $\lambda$ and using hypothesis iii we have that first two term can be ignored (don't forget that $r<\lambda$) and we can rearrange into $h_0 = \frac{2 \ell^2}{\lambda}$. Substituting $\ell$ and $\lambda$ we have Jurin law. Since the law states that $h_0$ grow when $R$ decrease, it will works for every smaller $R$ (every smaller $R$ too must satisfy condition iii). I underline that I exploited hypothesis iii, so this proof works only if the radius of the tube is much smaller than $h_0 \cos \theta$ and its square is much smaller than $2 \ell^2 \cos^2 \theta$: if contact angle is not small these could be, in practice, problematic conditions (unless it is possible to find a more generic proof).

• In Cengel book I found a simpler proof. The weight of the column is approximately $\pi R^2 h_0 \rho g$, and it is carried by the net surface tension force on the border: $2\pi R \sigma \cos \theta$. Equaling we can rearrange into Jurin law. But still there is the problem of sufficient and necessary conditions under which Jurin law works. Surely the approximated assumption about column weight works if $h(R)-h_0 \ll h_0$ (this is a sufficient condition) but this is not very satisfactory: what about large tubes? It is not clear to me if Jurin law works in general or works only for small tubes. – Fausto Vezzaro Mar 4 '15 at 21:28