Fluid Mechanics Concept Page - 18

Definition

Shape of liquid drop based on surface tension

T1=$T_1=$ Force due to surface tension at the liquid-solid interface
T2=$T_2=$ Force due to surface tension at the air-solid interface
T3=$T_3=$ Force due to surface tension at the air-liquid interface
For the equilibrium of the drop,
T2=T1+T3cosθ$T_2=T_1+T_3\cos \theta$
⟹cosθ=T2−T1T3$⟹\cos \theta ={\displaystyle \frac{T_2-T_1}{T_3}}$

Definition

Meniscus of a liquid

It is observed that when a free surface of a liquid comes in contact with a solid, it shows a curvy shape near the point of contact. Here an inward depression or an outward bulge takes place. This is called meniscus of a liquid.

Definition

Fluid flowing in a horizontal capillary

The factors on which fluid flowing in a horizontal capillary depends:
1) Pressure drop across the tube
2) Viscosity of the fluid
3) Diameter of the tube

Formula

Surface tension of water using capillary rise method

The surface tension of water can be calculated using capillary rise method.
The experiment can be conducted as follows:

• Take a clean glass tube. The radius of the glass tube should be small (of the order a millimeter). Let the radius of the tube be a$a$.
• Fill a beaker with water. insert the glass tube in the beaker vertically (as shown in the figure). Water will rise in the beaker. Let the height of water risen be h$h$.
• We know the angle of contact for the glass tube is 0o$0^o$.
• We also know that h=2Scos(θ)ρga$h={\displaystyle \frac{2Scos(\theta )}{\rho ga}}$, where S is surface tension of water.
• Put θ=0o$\theta =0^o$ and rearrange the formula. Then we get, surface tension S=ρgha2$S={\displaystyle \frac{\rho gha}{2}}$.
• Putting the known for water we can calculate the surface tension of water.

Definition

Capillarity and its Physical Example

 Capillarity, rise or depression of a liquid in a small passage such as a tube of small cross-sectional area, like the spaces between the fibres of a towel or the openings in a porous material. Capillarity is not limited to the vertical direction. Water is drawn into the fibres of a towel, no matter how the towel is oriented.

Liquids that rise in small-bore tubes inserted into the liquid are said to wet the tube, whereas liquids that are depressed within thin tubes below the surface of the surrounding liquid do not wet the tube. Water is a liquid that wets glass capillary tubes; mercury is one that does not. When wetting does not occur, capillarity does not occur.

Capillarity is the result of surface, or interfacial, forces. The rise of water in a thin tube inserted in water is caused by forces of attraction between the molecules of water and the glass walls and among the molecules of water themselves. These attractive forces just balance the force of gravity of the column of water that has risen to a characteristic height. The narrower the bore of the capillary tube, the higher the water rises. Mercury, conversely, is depressed to a greater degree, the narrower the bore.

Example

Problems on contact angle

Example: Two different vertical positions (a) and (b) of a capillary tube are shown in the figure with the lower end inside water. For position (a) Contact angle is 45o${45}^o$ and water rises to height h above the surface of water, while for position (b) height of the tube outside water is kept insufficient and equal to h2√${\displaystyle \frac{h}{\sqrt{2}}}$. Then what is the contact angle?

Solution:
h=2Tcos45∘rpg−−−−−(i)$h={\displaystyle \frac{2T\cos {45}^{\circ }}{rpg}}-----(i)$
h2√=2Tcosθrpg−−−−−(ii)${\displaystyle \frac{h}{\sqrt{2}}}={\displaystyle \frac{2T\cos \theta }{rpg}}-----(ii)$
(i)$(i)$ & (ii)⇒cosθ=12⇒θ=60∘$(ii)\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$

Example

Rise in capillary tube using force balance

The rise of a column of liquid within a fine capillary tube is also due to surface tension. Capillary action causes liquid to soak upwards through a piece of blotting paper and it also partly explains the rise of water through the capillaries in the stems of plants. Let the radius of the glass capillary tube be r, the coefficient of surface tension of the liquid he T, the density of the liquid be , the angle of contact between the liquid and the walls of the tube be and the height to which the liquid rises in the tube be h.  Consider the circumference of the liquid surface where it meets the glass. Along this line the vertical component of the surface tension force will be 2rcosθT$2rcos\theta T$.

Therefore
2rcosθT=r2gh$2rcos\theta T=r^{2}gh$
which gives
Capillary rise (h)=[2Tcosθ]/[rg]$(h)=[2Tcos\theta ]/[rg]$

Example

Fall in capillary tube

Example:
A capillary tube of diameter 0.4 mm$0.4mm$ is dipped in a beaker containing mercury of density 13.6×10$13.6\times 10$3${}^3$ kgm$kgm$−3${}^{-3}$ and surface tension 0.49 Nm$0.49Nm$−1${}^{-1}$.The angle of contact of mercury w.r.t. glass is 130$130$o${}^o$. [cos 130$cos130$o=${}^o=$ −0.64280$-0.64280$]. Find the depression of the meniscus in the capillary tube (g =$=$ 9.8 ms$9.8ms$−2${}^{-2}$).Solution:
The depression of meniscus in the capillary tube is 
= 2Tcosθρgr${\displaystyle \frac{2Tcos\theta }{\rho gr}}$
= 2×0.49×0.6428013.6×103×0.42×10−3×9.8${\displaystyle \frac{2\times 0.49\times 0.64280}{13.6\times {10}^3\times {\displaystyle \frac{0.4}{2}}\times {10}^{-3}\times 9.8}}$
= 0.024 m$0.024m$
=2.4 cm$2.4cm$

Definition

Shape of interface in a capillary tube of insufficient length

Rise in a capillary tube is given by: h=2TRg$h={\displaystyle \frac{2T}{Rg}}$
When length of the tube (l$l$) is smaller than h, lR′=2Tg$l{R}^{\prime }={\displaystyle \frac{2T}{g}}$
The value of R′$R^{\prime }$ increases until the relation is satisfied.
Shape of interface:
A liquid rises up into a capillary tube, dipped into it, until the weight of the liquid in the tube is just balanced by the force due to its surface tension. If q be the angle of contact between the liquid and the tube, and R$R$, the radius of liquid meniscus in the tube, we have r=Rcosq$r=R\cos q$, where r is the radius of the tube; so that,
T=Rcosθhρg2cosθ=hρgR2$T={\displaystyle \frac{R\cos \theta h\rho g}{2\cos \theta }}={\displaystyle \frac{h\rho gR}{2}}$
where h$h$ is the height of the liquid column in the tube.
Here clearly,
Rh=2Tρg$Rh={\displaystyle \frac{2T}{\rho g}}$Now with the tube sufficiently longer than h$h$, it is the value of h$h$ alone that changes to satisfy the above relation for T$T$. But if the tube be smaller than the calculated value of h$h$, the only variable in the above relation is R$R$, because now h=l$h=l$, the length of the tube (a constant) and so is q$q$ a constant for the given liquid and the tube. The liquid thus just spreads over the walls of the tube at the top and its meniscus acquires a new radius of curvature R′$R^{\prime }$, such that R′l=2Tρg$R^{\prime }l={\displaystyle \frac{2T}{\rho g}}$, or that R′l=R.h=$R^{\prime }l=R.h=$ a constant. And since l$l$ is smaller than R′>R$R^{\prime }>R$, i.e., the meniscus becomes less curved.

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