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-   -   About Young-laplacian equation. (http://www.cfd-online.com/Forums/main/119307-about-young-laplacian-equation.html)

sharonyue June 13, 2013 20:30

About Young-laplacian equation.
 
Hi guys,

In physics, the Young–Laplace equation is a nonlinear partial differential equation that describes the capillary pressure difference sustained across the interface between two static fluids, such as water and air, due to the phenomenon of surface tension. But how can I deduce this equation?http://en.wikipedia.org/wiki/Young%E...place_equation

\Delta p=- \gamma \nabla n

\kappa = - \nabla \cdot \left(\frac{\nabla \alpha}{|\nabla \alpha |}\right)

michujo June 14, 2013 03:54

Hi, just take a differential shell element (which represents the interface between both fluids) with different radius of curvature on each side (R1 and R2). The length of the element sides are dx1 and dx2, which span a differential angle d\theta_1 and d\theta_2 (where dx_1=R_1\cdot d\theta_1 and dx_2=R_2\cdot d\theta_2).

Consider static equilibrium of the shell element and you get something like this:

\Delta p=\gamma \cdot \left[ 2\cdot dx_2 \cdot sin( d\theta_1 /2)+2\cdot dx_1 \cdot sin( d\theta_2 /2) \right]

Assume small angles (sin(theta) approx. theta) and you get this:

\Delta p=\gamma \cdot \left( 1/R_1+1/R_2\right )

The expression comes from assuming that the pressure difference (which pushes the shell upwards normally to its surface, considering the higher pressure on the concave side of the shell) balances the forces due to surface tension, which pull the shell downwards at angles d\theta_1 /2 and d\theta_2 /2.

The easiest way is to draw the shell element with the dimensions and the force vectors and deducing the equilibrium equation.

Cheers,
Michujo.

sharonyue June 14, 2013 04:53

Quote:

Originally Posted by michujo (Post 433961)
Hi, just take a differential shell element (which represents the interface between both fluids) with different radius of curvature on each side (R1 and R2). The length of the element sides are dx1 and dx2, which span a differential angle d\theta_1 and d\theta_2 (where dx_1=R_1\cdot d\theta_1 and dx_2=R_2\cdot d\theta_2).

Consider static equilibrium of the shell element and you get something like this:

\Delta p=\gamma \cdot \left[ 2\cdot dx_2 \cdot sin( d\theta_1 /2)+2\cdot dx_1 \cdot sin( d\theta_2 /2) \right]

Assume small angles (sin(theta) approx. theta) and you get this:

\Delta p=\gamma \cdot \left( 1/R_1+1/R_2\right )

The expression comes from assuming that the pressure difference (which pushes the shell upwards normally to its surface, considering the higher pressure on the concave side of the shell) balances the forces due to surface tension, which pull the shell downwards at angles d\theta_1 /2 and d\theta_2 /2.

The easiest way is to draw the shell element with the dimensions and the force vectors and deducing the equilibrium equation.

Cheers,
Michujo.

Hi Michujo,
Thanks alot, after draw the shell element, I know \Delta p=\gamma \cdot \left( 1/R_1+1/R_2\right ), but how can I turn it to \Delta p=\gamma \cdot \nabla n?
Now I only know n=\nabla \alpha

Could it be said that \left( 1/R_1+1/R_2\right ) =- \nabla n?

michujo June 14, 2013 07:53

Hi, follow the variation of the normal vector along the element. Imagine that, at the center of the shell element, the normal is point upwards but, if you move along one side in the, let's say, the x direction, the normal at the edge of the element will be crooked because of the curvature of the element.
At the edge the normal will not be pointing exactly upwards but somewhat along the x direction. How much is "somewhat"? => sin(d\theta_1/2) \approx d\theta_1/2. The variation in the x coordinate is dx/2 (half length of the element along the x coordinate).

Therefore the variation of the x component of the normal vector as we move along the x coordinate is: n_{x_1}|_{x_1+dx_1/2}-n_{x_1}|_{x_1}=\frac{\partial n_{x_1}}{\partial x_1}\cdot dx_1/2=d\theta_1/2 and re-arranging: \frac{\partial n_{x_1}}{\partial x_1}=d\theta_1/dx_1=1/R_1

Thus \nabla \cdot n= \frac{\partial n_{x_1}}{\partial x_1}+\frac{\partial n_{x_2}}{\partial x_2}=1/R_1+1/R_2 expresses the variation of the normal components due to curvature of the fluids interface.

It's just some geometry and some algebraic manipulations.

Cheers,
Michujo.

sharonyue June 14, 2013 08:26

Quote:

Originally Posted by michujo (Post 433992)
Hi, follow the variation of the normal vector along the element. Imagine that, at the center of the shell element, the normal is point upwards but, if you move along one side in the, let's say, the x direction, the normal at the edge of the element will be crooked because of the curvature of the element.
At the edge the normal will not be pointing exactly upwards but somewhat along the x direction. How much is "somewhat"? => sin(\theta_1/2) \approx \theta_1/2. The variation in the x coordinate is dx/2 (half length of the element along the x coordinate).

Therefore the variation of the x component of the normal vector as we move along the x coordinate is: n_{x_1}|_{x_1+dx_1/2}-n_{x_1}|_{x_1}=\frac{\partial n_{x_1}}{\partial x_1}\cdot dx_1/2=\theta_1/2 and re-arranging: \frac{\partial n_{x_1}}{\partial x_1}=d\theta_1/dx_1=1/R_1

Thus \nabla \cdot n= \frac{\partial n_{x_1}}{\partial x_1}+\frac{\partial n_{x_2}}{\partial x_2}=1/R_1+1/R_2 expresses the variation of the normal components due to curvature of the fluids interface.

It's just some geometry and some algebraic manipulations.

Cheers,
Michujo.

Hi,

I just came back. Thats beautiful!! I need time to see it deeply, Thanks very much Michujo!!!

michujo June 14, 2013 08:53

1 Attachment(s)
No worries. I drew it so it's much easier to see instead of talking about vectors and differentials...


Cheers,
Michujo.

sharonyue June 14, 2013 09:47

1 Attachment(s)
Quote:

Originally Posted by michujo (Post 434002)
At the edge the normal will not be pointing exactly upwards but somewhat along the x direction. How much is "somewhat"? => .

Hi Michujo,

I just think about it deeply, Thats more clear now. Thats very kool. But there is only one thing left now: I dont understand why the increment in x-component of the normal vector equals sin d theta/2, could you explain it for a little bit? Thanks for your patience!!Or you can recommend me some papers about this.

In this image, q's magnitude equals dtheta/2?

michujo June 15, 2013 10:53

Hi, yes, in the image the magnitude of q is d\theta/2.

It's all geometry:
1) The interface element spans d\theta_1/2 and d\theta_2/2 along the x1 and x2 coordinates.

2) Also, you know that the modulus of the normal vector is 1.

3) Finally, as the normal vector is always perpendicular to the interface, by geometry, as the element turns and angle of d\theta/2 so will the normal vector. Therefore the components of the normal vector will be n_{x1}=1\cdot sin(d\theta_1/2) and n_{z}=1\cdot cos(d\theta_2/2). For that you have moved a distance dx1/2 from the center of the element.

4) sin(\theta) tends to \theta as \theta tends to zero.

I cannot think of any reference now but I'm sure there must be tons of information out there somewhere.

Cheers,
Michujo.

sharonyue June 15, 2013 19:26

Quote:

Originally Posted by michujo (Post 434142)
Hi, yes, in the image the magnitude of q is d\theta/2.

It's all geometry:
1) The interface element spans d\theta_1/2 and d\theta_2/2 along the x1 and x2 coordinates.

2) Also, you know that the modulus of the normal vector is 1.

3) Finally, as the normal vector is always perpendicular to the interface, by geometry, as the element turns and angle of d\theta/2 so will the normal vector. Therefore the components of the normal vector will be n_{x1}=1\cdot sin(d\theta_1/2) and n_{z}=1\cdot cos(d\theta_2/2). For that you have moved a distance dx1/2 from the center of the element.

4) sin(\theta) tends to \theta as \theta tends to zero.

I cannot think of any reference now but I'm sure there must be tons of information out there somewhere.

Cheers,
Michujo.

Woo, I figure this out totally.! Thanks buddy.!

Cheers!


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