# Page:EB1911 - Volume 05.djvu/286

273
CAPILLARY ACTION

The criterion as to whether any given catenoid is stable or not may be obtained as follows:—

Fig. 14.

Let ${\displaystyle {\mbox{PABQ}}}$ and ${\displaystyle {\mbox{A}}pq{\mbox{B}}}$ (fig. 14) be two catenaries having the same directrix and intersecting in ${\displaystyle A}$ and ${\displaystyle B}$. Draw ${\displaystyle {\mbox{P}}p}$ and ${\displaystyle {\mbox{Q}}q}$ touching both catenaries, ${\displaystyle {\mbox{P}}p}$ and ${\displaystyle {\mbox{Q}}q}$ will intersect at T, a point in the directrix; for since any catenary with its directrix is a similar figure to any other catenary with its directrix, if the directrix of the one coincides with that of the other the centre of similitude must lie on the common directrix. Also, since the curves at ${\displaystyle {\mbox{P}}}$ and ${\displaystyle p}$ are equally inclined to the directrix, ${\displaystyle {\mbox{P}}}$ and ${\displaystyle p}$ are corresponding points and the line ${\displaystyle {\mbox{P}}\;p}$ must pass through the centre of similitude. Similarly ${\displaystyle {\mbox{Q}}q}$ must pass through the centre of similitude. Hence ${\displaystyle {\mbox{T}}}$, the point of intersection of ${\displaystyle {\mbox{P}}p}$ and ${\displaystyle {\mbox{Q}}q}$, must be the centre of similitude and must be on the common directrix. Hence the tangents at A and B to the upper catenary must intersect above the directrix, and the tangents at A and B to the lower catenary must intersect below the directrix. The condition of stability of a catenoid is therefore that the tangents at the extremities of its generating catenary must intersect before they reach the directrix.

Stability of a Plane Surface.—We shall next consider the limiting conditions of stability of the horizontal surface which separates a heavier fluid above from a lighter fluid below. Thus, in an experiment of F. Duprez (“Sur un cas particulier de l’équilibre des liquides,” Nouveaux Mém. del’ Acad. de Belgique, 1851 et 1853), a vessel containing olive oil is placed with its mouth downwards in a vessel containing a mixture of alcohol and water, the mixture being denser than the oil. The surface of separation is in this case horizontal and stable, so that the equilibrium is established of itself. Alcohol is then added very gradually to the mixture till it becomes lighter than the oil. The equilibrium of the fluids would now be unstable if it were not for the tension of the surface which separates them, and which, when the orifice of the vessel is not too large, continues to preserve the stability of the equilibrium.

When the equilibrium at last becomes unstable, the destruction of equilibrium takes place by the lighter fluid ascending in one part of the orifice and the heavier descending in the other. Hence the displacement of the surface to which we must direct our attention is one which does not alter the volume of the liquid in the vessel, and which therefore is upward in one part of the surface and downward in another. The simplest case is that of a rectangular orifice in a horizontal plane, the sides being ${\displaystyle a}$ and ${\displaystyle b}$.

Let the surface of separation be originally in the plane of the orifice, and let the co-ordinates ${\displaystyle x}$ and ${\displaystyle y}$ be measured from one corner parallel to the sides ${\displaystyle a}$ and ${\displaystyle b}$ respectively, and let ${\displaystyle z}$ be measured upwards. Then if ${\displaystyle \rho }$ be the density of the upper liquid, and ${\displaystyle \sigma }$ that of the lower liquid, and ${\displaystyle {\mbox{P}}}$ the original pressure at the surface of separation, then when the surface receives an upward displacement ${\displaystyle z}$, the pressure above it will be ${\displaystyle {\mbox{P}}-\rho gz}$, and that below it will be ${\displaystyle {\mbox{P}}-\rho gz}$, so that the surface will be acted on by an upward pressure ${\displaystyle (\rho -\sigma )gz}$. Now if the displacement ${\displaystyle z}$ be everywhere very small, the curvature in the planes parallel to ${\displaystyle xz}$ and ${\displaystyle yz}$ will be ${\displaystyle d^{2}z/dx^{2}}$ and ${\displaystyle d^{2}z/dy^{2}}$ respectively, and if ${\displaystyle {\mbox{T}}}$ is the surface-tension the whole upward force will be

${\displaystyle {\mbox{T}}\left({\frac {d^{2}z}{dx^{2}}}+{\frac {d^{2}z}{dy^{2}}}\right)+(\rho -\sigma )gz.}$

If this quantity is of the same sign as ${\displaystyle z}$, the displacement will be increased, and the equilibrium will be unstable. If it is of the opposite sign from ${\displaystyle z}$, the equilibrium will be stable. The limiting condition may be found by putting it equal to zero. One form of the solution of the equation, and that which is applicable to the case of a rectangular orifice, is

${\displaystyle z={\mbox{C}}\sin px\sin qy.}$

Substituting in the equation we find the condition

${\displaystyle (p^{2}+q^{2}){\mbox{T}}-(\rho -\sigma )g={\begin{cases}+^{ve}&{\mbox{stable.}}\\{\mbox{o}}&{\mbox{neutral.}}\\-^{ve}&{\mbox{unstable.}}\end{cases}}}$

That the surface may coincide with the edge of the orifice, which is a rectangle, whose sides are a and b, we must have

${\displaystyle pa=m\pi \quad ,\;qb=n\pi ,}$

when ${\displaystyle m}$ and ${\displaystyle n}$ are integral numbers. Also, if ${\displaystyle m}$ and ${\displaystyle n}$ are both unity, the displacement will be entirely positive, and the volume of the liquid will not be constant. That the volume may be constant, either ${\displaystyle n}$ or ${\displaystyle m}$ must be an even number. We have, therefore, to consider the conditions under which

${\displaystyle \pi ^{2}\left({\frac {m^{2}}{a^{2}}}+{\frac {n^{2}}{b^{2}}}\right){\mbox{T}}-(\rho -\sigma )g}$

cannot be made negative. Under these conditions the equilibrium is stable for all small displacements of the surface. The smallest admissible value of ${\displaystyle {\tfrac {m^{2}}{a^{2}}}+{\tfrac {n^{2}}{b^{2}}}}$ is ${\displaystyle {\tfrac {4}{a^{2}}}+{\tfrac {1}{b^{2}}}}$, where ${\displaystyle a}$ is the longer side of the rectangle. Hence the condition of stability is that

${\displaystyle \pi ^{2}\left({\frac {4}{a^{2}}}+{\frac {1}{b^{2}}}\right){\mbox{T}}-(\rho -\sigma )g}$

is a positive quantity. When the breadth ${\displaystyle b}$ is less than ${\displaystyle {\sqrt {\frac {\pi ^{2}{\mbox{T}}}{(\rho -\sigma )g}}}}$ the length ${\displaystyle a}$ may be unlimited.

When the orifice is circular of radius ${\displaystyle a}$, the limiting value of ${\displaystyle a}$ is ${\displaystyle {\sqrt {\frac {\mbox{T}}{g\rho z}}}}$, where ${\displaystyle z}$ is the least root of the equation

${\displaystyle {\frac {2}{z}}{\mbox{J}}_{1}(z)=1-{\frac {z^{2}}{2\cdot 4}}+{\frac {z^{4}}{2\cdot 4^{2}\cdot 6}}-{\frac {z^{6}}{2\cdot 4^{2}\cdot 6^{2}\cdot 8}}+\&{\mbox{c.,}}=0}$

The least root of this equation is

${\displaystyle z=3.83171}$.

If ${\displaystyle h}$ is the height to which the liquid will rise in a capillary tube of unit radius, then the diameter of the largest orifice is

${\displaystyle 2a=3.83171{\sqrt {(2h)}}=5.4188{\sqrt {(h)}}.}$

Duprez found from his experiments

${\displaystyle 2a=5.485{\sqrt {(h)}}.}$

[The above theory may be well illustrated by a lecture experiment. A thin-walled glass tube of internal diameter equal to 14½ mm. is ground true at the lower end. The upper end is contracted and is fitted with a rubber tube under the control of a pinch-cock. Water is sucked up from a vessel of moderate size, the rubber is nipped, and by a quick motion the tube and vessel are separated, preferably by a downward movement of the latter. The inverted tube, with its suspended water, being held in a clamp, a beaker containing a few drops of ether is brought up from below until the free surface of the water is in contact with ether vapour. The lowering of tension, which follows the condensation of the vapour, is then strikingly shown by the sudden precipitation of the water.]

Effect of Surface-tension on the Velocity of Waves.—When a series of waves is propagated on the surface of a liquid, the surface-tension has the effect of increasing the pressure at the crests of the waves and diminishing it in the troughs. If the wave-length is λ, the equation of the surface is

${\displaystyle y=b\sin 2\pi {\frac {x}{\lambda }}}$

The pressure due to the surface tension T is

${\displaystyle p=-{\mbox{T}}{\frac {d^{2}y}{dx^{2}}}={\frac {4\pi ^{2}}{\lambda ^{2}}}{\mbox{T}}y}$

This pressure must be added to the pressure due to gravity g${\displaystyle \rho }$y. Hence the waves will be propagated as if the intensity of gravity had been

${\displaystyle f=g+{\frac {4\pi ^{2}}{\lambda ^{2}}}{\frac {\mbox{T}}{\rho }}}$

instead of ${\displaystyle g}$. Now it is shown in hydrodynamics that the velocity of propagation of waves in deep water is that acquired by a heavy body falling through half the radius of the circle whose circumference is the wave-length, or

 ${\displaystyle v^{2}={\frac {f\lambda }{2\pi }}={\frac {g\lambda }{2\pi }}+{\frac {2\pi {\mbox{T}}}{\rho \lambda }}}$ (1)

This velocity is a minimum when

${\displaystyle \lambda =2\pi {\sqrt {\frac {\mbox{T}}{g\rho '}}}}$

and the minimum value is

${\displaystyle v={\sqrt[{4}]{4{\frac {{\mbox{T}}g.}{\rho }}}}}$

For waves whose length from crest to crest is greater than ${\displaystyle \lambda }$, the principal force concerned in the motion is that of gravitation.