About Young-laplacian equation.

sharonyue · June 13, 2013, 21:30

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, 04:54

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, 05:53

Quote:

Originally Posted by michujo

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, 08:53

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, 09:26

Quote:

Originally Posted by michujo

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, 09:53

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, 10:47

Quote:

Originally Posted by michujo

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

, 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

?

michujo · June 15, 2013, 11:53

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, 20:26

Quote:

Originally Posted by michujo

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!

June 13, 2013, 21:30	About Young-laplacian equation.	#1
sharonyue Senior Member Dongyue Li Join Date: Jun 2012 Location: Beijing, China Posts: 849 Rep Power: 18	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 $\Delta p=- \gamma \nabla n$ $\kappa = - \nabla \cdot \left(\frac{\nabla \alpha}{\|\nabla \alpha \|}\right)$ Last edited by sharonyue; June 14, 2013 at 04:06.

June 14, 2013, 08:53		#4
michujo Senior Member Join Date: Dec 2011 Location: Madrid, Spain Posts: 134 Rep Power: 16	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 likes this. Last edited by michujo; June 14, 2013 at 10:00.

June 15, 2013, 11:53		#8
michujo Senior Member Join Date: Dec 2011 Location: Madrid, Spain Posts: 134 Rep Power: 16	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 likes this.

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June 14, 2013, 04:54		#2
michujo Senior Member Join Date: Dec 2011 Location: Madrid, Spain Posts: 134 Rep Power: 16	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.