# Evaporation from a capillary tube

Consider a capillary tube (say from a liquid / capillary thermometer), that means a tube of small internal diameter which holds liquid by capillary action . The tube is fulfilled with water and closed at one end. How long it would take for water to completely evaporate out through the open end of the tube?

To be specific, let the diameter of the tube be $d=1$mm, the length of the tube $L=1$m, the environmental temperature $T=300$ K and pressure $P=10^5$ Pa, the relative humidity $f=60$ %, the contact angle a) $\theta=0^{\circ}$ and b) $\theta=180^{\circ}$.

Does it have any effect on the time when the tube is positioned horizontally or vertically(open end above)?

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"Do my homework"-type questions are not allowed here, and this looks very much like homework. See physics.stackexchange.com/faq#questions – Christoph Jun 22 '11 at 8:48
@user599884: I agree that it is formulated like a homework but only for that to be as specific as possible. If you're immersed into this problem in depth then i do not think this is a trivial homework problem. This problem is inspired by real life. – Martin Gales Jun 22 '11 at 9:09
In that case disregard what I said. – Christoph Jun 22 '11 at 9:16

As a rough answer: you can calculate the rate of evaporation from a water surface for the given temperature, humidity and a parabolic surface area.

The problem starts a few seconds later: The relative humidity above the water surface will rise close to 100% and now the rate of evaporation is limited by the diffusion of water molecules through the air out of the capillary. If you are not interested in the time dependence one can consider these two cases. After some time the rate of evaporation will be equal to the diffusion rate at the end of the capillary.

Additionally the length of the tube compared to the length of the water column can influence the result because it determines the length of the capillary through which the water vapour has to diffuse.

To your last question: In theory yes, as you will have less diffusion against the additional gravity, in practice it will only matter if you have a much longer tube. (The opposite is true if the experiment is surrounded by normal air. The mass of air molecules like O$_2$, N$_2$, ... is higher than H$_2$O as Georg pointed out)

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""as you will have less diffusion against the additional gravity"" As water vapour/moist air/water molecules are lighter than air, gravity will heĺp! Last (if tiny) effect to mention: vapor pressure of the water column is lowered by the extra (negative) pressure of capillary action. – Georg Nov 2 '11 at 11:57
@Georg: Oh, I did not consider the mass of the other gas molecules. At some point the H2O molecules will form bigger conglomerates and reverse the situation, but I guess we won't have clouds in the capillary. It is really intriguing how difficult it can get if you want to incorporate many factors. – Alexander Nov 2 '11 at 14:00
@Alexander: Thank you for you input. How about make a quick guess: What could be the evaporation time? One month, half years, year or maybe few years? – Martin Gales Nov 3 '11 at 6:19
@MartinGales: I would guess a year. At the moment I don't have a 1 mm capillary here but if the answer is connected to a real life problem it should not be hard to do the experiment. – Alexander Nov 3 '11 at 12:34

OK, i tried to do a rough estimation, supposing that the contact angle is $\theta=\frac{\pi}{2}$

A start point is Fick's law of diffusion

$$j=-D\frac{dn}{dx}$$ where $D$ $(\frac{m^2}{s})$ is diffusivity of water vapor in air , $n$ $(\frac{1}{m^3})$ is concentration of water vapor in air and $x$- axis lies in the center of the capillary. Using ideal gas law $p=nkT$

$$j=-\frac{D}{kT}\frac{dp}{dx}$$ If we multiply the last eq. with $m_0$(mass of water molecule) then

$$\frac{dm}{Sdt}=-\frac{D\mu}{RT}\frac{dp}{dx}$$ where $\mu$ is the molar mass, $R$ is the universal gas constant, $S$ is cross-sectional area of the capillary.
On the other hand, $m=\rho_w Sx$ where $x$ is the length of the water column in the capillary, $\rho_w$ is density of water. So

$$\frac{dx}{dt}=-\frac{D\mu}{\rho_w RT}\frac{dp}{dx}=-\frac{D}{\rho_w}\frac{d\rho}{dx}$$

Because the diffusion process is quasistatic, $\frac{d\rho}{dx}=\frac{\rho_s-\rho}{L-x}$ holds.

Here $\rho_s$ is saturation vapor density over flat water, $\rho=f\rho_s$ is water vapor density in the environment. Finally:

$$\frac{dx}{dt}=-\frac{D\rho_s(1-f)}{\rho_w(L-x)}$$

After integrating from $x=0$ to $x=L$ one gets evaporation time:

$$t=\frac{\rho_w L^2}{2D\rho_s(1-f)}$$

Now, a numerical estimation:

$L=1m$
$D=3*10^{-5}\frac{m^2}{s}$
$\rho_w=10^{3}\frac{kg}{m^3}$
$\rho_s=1.8*10^{-2}\frac{kg}{m^3}$
$f=0.6$

So $t=7.3$ years

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60% humidity for the environment? That's a bit higher than I would have used, but very nice answer. +1 – Rick Jan 22 '15 at 19:50