Jurin's law, orcapillary rise, is the simplest analysis ofcapillary action—the induced motion of liquids in small channels^[1]—and states that the maximum height of a liquid in a capillary tube isinversely proportional to the tube'sdiameter. Capillary action is one of the most common fluid mechanical effects explored in the field ofmicrofluidics. Jurin's law is named afterJames Jurin, who discovered it between 1718 and 1719.^[2] His quantitative law suggests that the maximum height of liquid in a capillary tube is inversely proportional to the tube's diameter. The difference in height between the surroundings of the tube and the inside, as well as the shape of themeniscus, are caused bycapillary action. The mathematical expression of this law can be derived directly fromhydrostatic principles and theYoung–Laplace equation. Jurin's law allows the measurement of thesurface tension of a liquid and can be used to derive thecapillary length.^[3]

The law is expressed as^{[citation needed]}

${\displaystyle h=\frac{2\gamma \cos \theta }{\rho g{r}_0}}$,

where

• h is the liquid height;
• ${\displaystyle \gamma }$ is thesurface tension;
• θ is thecontact angle of the liquid on the tube wall;
• ρ is the massdensity (mass per unit volume);
• r₀ is the tube radius;
• g is thegravitational acceleration.

It is only valid if the tube is cylindrical and has a radius (r₀) smaller than thecapillary length (${\displaystyle {\lambda }_{\mathrm{c}}^2=\gamma /(\rho g)}$). In terms of the capillary length, the law can be written as

${\displaystyle {\lambda }_{\mathrm{c}}^2=\frac{h{r}_0}{2\cos \theta }}$.

For a water-filled glass tube in air atstandard conditions for temperature and pressure,γ = 0.0728 N/m at 20 °C,ρ = 1000 kg/m³, andg = 9.81 m/s². Because waterspreads on clean glass, the effective equilibrium contact angle is approximately zero.^[4] For these values, the height of the water column is

${\displaystyle h\approx \frac{1.48\times {10}^{-5}}{r_0}\text{m}.}$

Thus for a 2 m (6.6 ft) radius glass tube in lab conditions given above, the water would rise an unnoticeable 0.007 mm (0.00028 in). However, for a 2 cm (0.79 in) radius tube, the water would rise 0.7 mm (0.028 in), and for a 0.2 mm (0.0079 in) radius tube, the water would rise 70 mm (2.8 in).

Capillary action is used by many plants to bring up water from the soil. For tall trees (larger than ~10 m (32 ft)), other processes likeosmotic pressure andnegative pressures are also important.^[5]

During the 15th century,Leonardo da Vinci was one of the first to propose thatmountain streams could result from the rise of water through small capillary cracks.^[3][6]

It is later, in the 17th century, that the theories about the origin of capillary action begin to appear.Jacques Rohault erroneously supposed that the rise of the liquid in a capillary could be due to the suppression of air inside and the creation of a vacuum. The astronomerGeminiano Montanari was one of the first to compare the capillary action to the circulation ofsap in plants. Additionally, the experiments ofGiovanni Alfonso Borelli determined in 1670 that the height of the rise was inversely proportional to the radius of the tube.

Francis Hauksbee, in 1713, refuted the theory of Rohault through a series of experiments on capillary action, a phenomenon that was observable in air as well as in vacuum. Hauksbee also demonstrated that the liquid rise appeared on different geometries (not only circular cross sections), and on different liquids and tube materials, and showed that there was no dependence on the thickness of the tube walls.Isaac Newton reported the experiments of Hauskbee in his workOpticks but without attribution.^[3][6]

It was the English physiologistJames Jurin, who finally in 1718^[2] confirmed the experiments of Borelli and the law was named in his honour.^[3][6]

The height${\displaystyle h}$ of the liquid column in the tube is constrained by thehydrostatic pressure and by thesurface tension. The following derivation is for a liquid that rises in the tube; for the opposite case when the liquid is below the reference level, the derivation is analogous but pressure differences may change sign.^[1]

Above the interface between the liquid and the surface, the pressure is equal to the atmospheric pressure${\displaystyle p_{\mathrm{a}\mathrm{t}\mathrm{m}}}$. At the meniscus interface, due to the surface tension, there is a pressure difference of${\displaystyle \mathrm{\Delta }p=p_{\mathrm{a}\mathrm{t}\mathrm{m}}-p_{\mathrm{i}\mathrm{n}\mathrm{t}}}$, where${\displaystyle p_{\mathrm{i}\mathrm{n}\mathrm{t}}}$ is the pressure on the convex side; and${\displaystyle \mathrm{\Delta }p}$ is known asLaplace pressure. If the tube has a circular section of radius${\displaystyle r_0}$, and the meniscus has a spherical shape, the radius of curvature is${\displaystyle r=r_0/\cos \theta }$, where${\displaystyle \theta }$ is thecontact angle. The Laplace pressure is then calculated according to theYoung-Laplace equation:$${\displaystyle \mathrm{\Delta }p=\frac{2\gamma }{r},}$$where${\displaystyle \gamma }$ is the surface tension.

Outside and far from the tube, the liquid reaches a ground level in contact with the atmosphere. Liquids incommunicating vessels have the same pressures at the same heights, so a point${\displaystyle \mathrm{w}}$, inside the tube, at the same liquid level as outside, would have the same pressure${\displaystyle p_{\mathrm{w}}=p_{\mathrm{a}\mathrm{t}\mathrm{m}}}$. Yet the pressure at this point follows avertical pressure variation as

${\displaystyle p_{\mathrm{w}}=p_{\mathrm{i}\mathrm{n}\mathrm{t}}+\rho gh,}$

where${\displaystyle g}$ is thegravitational acceleration and${\displaystyle \rho }$ the density of the liquid. This equation means that the pressure at point${\displaystyle \mathrm{w}}$ is the pressure at the interface plus the pressure due to the weight of the liquid column of height${\displaystyle h}$. In this way, we can calculate the pressure at the convex interface$${\displaystyle p_{\mathrm{i}\mathrm{n}\mathrm{t}}=p_{\mathrm{w}}-\rho gh=p_{\mathrm{a}\mathrm{t}\mathrm{m}}-\rho gh.}$$

The hydrostatic analysis shows that${\displaystyle \mathrm{\Delta }p=\rho gh}$, combining this with the Laplace pressure calculation we have:$${\displaystyle \rho gh=\frac{2\gamma \cos \theta }{r_0},}$$solving for${\displaystyle h}$ returns Jurin's law.

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