# Lagrange-Young equation in Fluid Mechanics

1. Aug 27, 2013

### Nikitin

Young-Laplace equation in Fluid Mechanics

EDIT: I meant the Young-Laplace equation, of course, not the Young-Lagrange.. sorry!

Heya!

According to said equation, ΔP*area = γ*circumference, for an interface of spherical fluid-element.

Can I pls get some explanations?

1) Why is tension, or γ, purely a result of chemical forces? Isn't there a pressure parallel to the surface, just like there is a pressure inside the fluid-element? Because if there were, the equation should be ΔP*area + P*area = γ*circumference.

2) Is there a physical explanation for the formula? Why does γ=ΔP*area/circumference?

Last edited: Aug 27, 2013
2. Aug 27, 2013

### voko

I am not familiar with the name Lagrange-Young equation, but it seems to describe how surface tension is related to the pressure difference over the surface.

Then it makes perfect sense. The force of tension at the boundary of some surface segment must stretch it, and the total "stretching" force over the entire boundary must be balanced by the force due to the pressure difference over the area of the segment.

Think about a balloon made of rubber. Because the rubber is stretched, it presses against that content of the balloon, thus the internal pressure is higher than the external. Now cut a piece of the rubber from the balloon. It will be acted upon by the pressure difference, tending to separate it from the rest of the balloon, so in order to keep it in place, you must apply force to its edges.

3. Aug 27, 2013

### Nikitin

You're right to not be familiar with it, because it's called the Laplace-Young equation.. Sorry, I keep mixing lagrange and laplace. :P here's what I was talking about: http://en.wikipedia.org/wiki/Young–Laplace_equation

Anyway: I read the explanation offered by Cimbala & Cengel's fluid mechanics, and I think I got it now. Though I am unsure on what delta-P actually is: is it the difference between the pressure inside the fluid and outside, or is it the pressure-change inside the fluid?

Last edited: Aug 27, 2013
4. Aug 27, 2013

### voko

The referenced article explains what $\Delta p$ is.

5. Aug 27, 2013

### Staff: Mentor

Pressure only acts normal to surfaces. There is no shear component of pressure. Any way you orient an area element within a fluid, the pressure only acts perpendicular to the area element. In a fluid, the viscous portion of the stress tensor has shear components on arbitrarily oriented area elements, but not the pressure portion.

Chet

6. Aug 28, 2013

### Andy Resnick

The second question is easier to answer: it arises from energy minimization of a dividing surface. The detailed calculation can be difficult to follow, but a good reference is here:

http://www.sciencedirect.com/science/article/pii/0001868695002820#

Similarly, Young's equation is derived from conservation of momentum at the three-phase line.

I may not understand what you are asking in question (1): the interfacial energy is just that- the energy associated with an interface. Since fluids can't support a shear stress, there is no 'parallel' component. Solids can support a shear, so the interfacial energy of a solid-vacuum interface is not easy to define (Wulff constructions are commonly used). A good starting point is here:

http://www.virginia.edu/ep/SurfaceScience/Thermodynamics.html

7. Aug 28, 2013

### Staff: Mentor

I think that this needs to be qualified a little. Fluids can't support a shear stress statically, but can exhibit shear stresses by deforming viscously with time (i.e., with time in a Lagrangian sense). For example, fluid being sheared continuously between parallel plates causes a shear stress on the plates.

Chet

8. Aug 28, 2013

### Andy Resnick

Yes, fluids flow in response to shear. Sometimes the flow is reversible, most often it is not. In any case, a fluid's inability to 'store' elastic energy is why there is no parallel component in the OP.

Edit: my comments above are restricted to Newtonian fluids only- viscoelastic fluids, bingham fluids, and other fluid phases possessing a yield stress violate what I said above.