Difference between revisions of "Surface curvature"

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(Simple derivation)
(Ostwald ripening)
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Ostwald ripening was first discovered in 1896 by Wilhem Ostwald. This process is driven by the fact that larger droplets are more energetically favorable.  This is because they are more thermordynamically favored whereas small droplets are more kinetically favored.  In an effort to reduce energy, states with the molecules well ordered and packed are more favorable.  The large volume to surface area ratio of large droplets allows for a lower energy state.  This will cause the smaller droplets to transform into large crystals.  (http://xray.bmc.uu.se/~terese/crystallization/tutorials/tutorial6.html)
  
 
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In this course, Ostwald ripening is significant because it is a key mechanism in the destabilization of emulsions (for example, by creaming and sedimentation).  Relating back to the previous posts about the importance of crystal formation in ice cream, Ostwald ripening can play a big role in making ice cream taste old and less creamy.  The water in the ice cream recrystallizes causing the larger crystals to grow competitively with the small crystals.  This destabilizes the homogeneity of the emulsion and leads to the gritty taste of older ice cream.  http://www.answers.com/topic/ostwald-ripening
 
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Revision as of 22:15, 1 October 2008

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Pierre Simon, Marquis de Laplace

Curvature of a liquid surface implies a pressure drop across that surface, the larger the curvature, the greater the pressure drop. The fundamental work was by Pierre Simon, Marquis de Laplace.

www.swlearning.com

Laplace equation

deGennes et al. give a simple (but not general) derivation for the relation between the surface tension, surface curvature, and pressure drop:

Overpressure inside a drop of oil “o” in water “w”. de Gennes (2004) Fig. 1.5
If the surface is perturbed from a sphere: <math>\delta W=-p_{o}dV_{o}-p_{w}dV_{w}+\sigma _{o/w}dA</math>
For a sphere: <math>\begin{align} & dV_{o}=4\pi R^{2}dR=-dV_{w} \\ & dA=8\pi RdR \\ \end{align}</math>
At equilibrium: <math>\begin{align} & \delta W=0 \\ & p_{o}-p_{w}=\Delta p=\frac{2\sigma _{o/w}}{R} \\ \end{align}</math>
For oil/water

<math>\Delta p=\frac{2\sigma _{o/w}}{R}\approx \frac{2\cdot 30\cdot 10^{-3}{N}/{m}\;}{R}</math>

For a 1mm drop: <math>\Delta p\approx \frac{6\cdot 10^{-2}{N}/{m}\;}{10^{-6}m}\approx 6\cdot 10^{4}Pa\approx 0.6\text{ atm}</math>






Ostwald ripening

For drops (or bubbles) of different sizes in the same emulsion (or foam), the internal pressures are different. The smaller is at higher pressure than the larger so that molecules diffuse from small drops to large ones. A process called Ostwald ripening.

BinanryBubbles.png

Ostwald ripening was first discovered in 1896 by Wilhem Ostwald. This process is driven by the fact that larger droplets are more energetically favorable. This is because they are more thermordynamically favored whereas small droplets are more kinetically favored. In an effort to reduce energy, states with the molecules well ordered and packed are more favorable. The large volume to surface area ratio of large droplets allows for a lower energy state. This will cause the smaller droplets to transform into large crystals. (http://xray.bmc.uu.se/~terese/crystallization/tutorials/tutorial6.html)

In this course, Ostwald ripening is significant because it is a key mechanism in the destabilization of emulsions (for example, by creaming and sedimentation). Relating back to the previous posts about the importance of crystal formation in ice cream, Ostwald ripening can play a big role in making ice cream taste old and less creamy. The water in the ice cream recrystallizes causing the larger crystals to grow competitively with the small crystals. This destabilizes the homogeneity of the emulsion and leads to the gritty taste of older ice cream. http://www.answers.com/topic/ostwald-ripening


General form of Laplace equation

A more general form of the Laplace equation is <math>\Delta p=\sigma \left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)</math> . Where R1 and R2 are the principle radii of curvature.

(This section needs (a) a derivation of the Laplace equation and (b) a description of the radii of curvature.)







Minimial surfaces

Bubble structures often imitate natural structures possible indicating that these shapes are controlled by the same forces that control foams:

(From lantern slide.)




Lord Kelvin

William Thomson, 1st Baron Kelvin, (26 June 1824 – 17 December 1907):

Weaire, 1996 Front Cover (in reverse?)

The Kelvin Problem

The Kelvin problem is that of partitioning 3D space into cells of equal volume and minimum surface area. Kelvin proposed the solution as a curved tetrakaidecahedron, six squares and eight hexagons. All edges are curved; all faces with no curvature, but none flat.


Weaire, p. 56

However, structure with even smaller surface areas have been found recently and are discussed in Weaire, 1996.



Bubble rise

The motion of bubbles is an active field of research. (A description of Prof. Stone's research should be added here.)


A model for the effect of viscosity on bubble formation has been given by Krotov (p. 203f).


Without viscous effects the pressure is determined by the Laplace equation: <math>p^{\alpha }\left( t \right)=p+\frac{2\sigma \left( t \right)}{R\left( t \right)}</math>

Adding the viscous effects gives: <math>p^{\alpha }\left( t \right)=p+\frac{2}{R\left( t \right)}\left( \sigma \left( t \right)+2\eta \frac{dR\left( t \right)}{dt} \right)</math>

The viscous term is negligible for aqueous systems but might be important for the maximum bubble pressure experiment or ink jetting. e.g.

<math>\begin{align}

 & \text{For glycerol:} \\ 
& \eta =1.49\text{ }Pa-s\text{  and with }\frac{dR\left( t \right)}{dt}\approx 10^{-3}\text{ }m-s \\ 
& 2\eta \frac{dR\left( t \right)}{dt}=3\text{ }{mN}/{m}\; \\ 

\end{align}</math>



Capillary adhesion

A liquid drop between two flat plates has curved surfaces, but with curvatures in opposite directions:

de Gennes, 2004, Fig. 1.8

The large radius of curvature, R, is in the plane of the plates. The smaller one is perpendicular and opposite in sign: <math>-{}^{H}\!\!\diagup\!\!{}_{\left( 2\cos \theta _{e} \right)}\;</math>

The Laplace pressure is: <math>\Delta p=\sigma \left( \frac{1}{R}-\frac{2\cos \theta _{e}}{H} \right)\sim -\frac{2\sigma \cos \theta _{e}}{H}</math>

The force pushing the plates together is the drop area times Laplace pressure: <math>F=\pi R^{2}\frac{2\sigma \cos \theta _{e}}{H}</math>

<math>\text{For }R=1\text{ cm, }H=5\mu \text{m, and }\theta =0</math> the pressure is about 1/3 atm and the force 10 N.


(What is the stable state? What are the dynamics?)

Drops on a fiber

For small drops, gravity is not significant so the pressure is constant throughout the drop, hence the curvature is constant everywhere:

de Gennes, 2004 Fig.1-10


<math>\begin{align}

 & \Delta p=\sigma \left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)\text{ }\Rightarrow \text{ }\frac{\Delta p}{\sigma }=\left( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right)=\text{constant} \\ 
& \text{ or  }\left( \frac{{\ddot{z}}}{\left[ 1+\dot{z}^{2} \right]^{{3}/{2}\;}}+\frac{{\dot{z}}}{x\left[ 1+\dot{z} \right]^{{1}/{2}\;}} \right)=\text{constant} \\ 

\end{align}</math>

de Gennes (2004, p. 11) gives the solution by considering the forces on the vertical dotted line. For any position on the x-axis, where the thickness of the drop is z, the sum of the component of surface tension pulling to the right and the force of pressure to the left and the force of wetting the fiber is zero: <math>2\pi z\sigma \cos \theta -\Delta p\pi \left( z^{2}-b^{2} \right)-2\pi b\sigma =0</math>

<math>\text{From p}\text{. 12 }\cos \theta ={1}/{\sqrt{1+\dot{z}^{2}}}\;</math>

The equation applies for all z, hence when <math>\dot{z}=0\text{ and }z=L</math>

Hence: <math>\Delta p=\frac{2\sigma }{L+b}</math>





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