Fluid Equilibrium Interface Shape

In summary, the conversation discusses the shape of liquid in a 2D rectangular channel in static equilibrium. The equations of motion and Laplace pressure jump are considered, and a pressure balance equation is derived. The conversation also mentions a technique of solving 5 linearized equations to determine the behavior with oscillations, and suggests resolving the linearized equation into harmonics to reduce it to a sequence of uncoupled linear ODEs. The conversation ends with a request for further elaboration or literature recommendation.
  • #1
member 428835
Hi PF!

Suppose a 2D rectangular channel with wall-normal perpendicular to the downward gravity force ##F## rests is static equilibrium. What shape would the liquid take?

Looking at the equations of motion, I think the following would have to be satisfied: $$\frac{d P}{dy} = -\rho F\\
\Delta P = \sigma k_1$$
1) Eulers equation for 1D (since gravity not in horizontal ##x##-direction)
2) Laplace pressure jump where ##\Delta P## is pressure jump from inside to outside liquid interface and 2D implies curvature ##k_2 = 0##, thus not listed.

Integrating Eulers eq yields ##P = -\rho F y + c##.

If we call the meniscus ##\Gamma(x)##, then it's curvature must be ##k_1(x) = \Gamma''(x) / (1+\Gamma(x)^2)^{3/2}##. Evaluating the pressure along the meniscus and applying the Laplace pressure jump implies $$\sigma \frac{\Gamma ''(x)}{(1+\Gamma'(x)^2 )^{3/2}} = \rho F \Gamma(x)$$ subject to $$\Gamma'(0) = -\cot \theta\\ \Gamma'(1) = 0$$ where I assume the channel has length 2. But how is this equation solved; the IC's are both at the first derivative.

For help purposes, seems like a good question for @Chestermiller and @boneh3ad though anyone's thoughts are welcomed.
 
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  • #2
I haven't rederived your equation to confirm that it is correct, but, of course, this 2nd order non-linear ODE with split boundary conditions can be solved numerically.
 
  • #3
Chestermiller said:
I haven't rederived your equation to confirm that it is correct, but, of course, this 2nd order non-linear ODE with split boundary conditions can be solved numerically.
Thanks for checking it out! I have a question for you; I need an analytic solution in order to examine small linear normal oscillations. Then would you suggest a good approach is to get the numerical solution and curve-fit it, and use that? Or are there more sophisticated techniques?

I noticed this problem is a well understood problem and admits a closed analytic solution if I use as a BC ##\Gamma(\infty) \to 0##. Which is close to zero at ##\Gamma'(1)##.
 
  • #4
joshmccraney said:
Suppose a 2D rectangular channel with wall-normal perpendicular to the downward gravity force ##F## rests is static equilibrium. What shape would the liquid take?

It's been a while since I worked on this kind of problem, but I suspect you are oversimplifying the system- for example, I'm not sure you can simply ignore the 3D aspect of the problem. I found some relevant results, hope these are helpful:

http://adsabs.harvard.edu/abs/2005PhyD..209..236Shttps://www.cambridge.org/core/jour...clined-plane/7AEF7AF42BC3311BB42A193FE6F44FE5https://pdfs.semanticscholar.org/59e8/b936dc64535a898064258dfb8bc052d01f7e.pdf
 
  • #5
joshmccraney said:
Thanks for checking it out! I have a question for you; I need an analytic solution in order to examine small linear normal oscillations. Then would you suggest a good approach is to get the numerical solution and curve-fit it, and use that? Or are there more sophisticated techniques?

I noticed this problem is a well understood problem and admits a closed analytic solution if I use as a BC ##\Gamma(\infty) \to 0##. Which is close to zero at ##\Gamma'(1)##.
I'm not familiar with that analytic solution, but even if you know that, how would you then include oscillations.

I would start out by writing the PDE for the behavior with oscillations, then linearize with respect to the oscillations, and then separate that into the non-linear equation you have already plus the linearized equation for the oscillations (based on the nonlinear static equation solution). Then I would resolve the linearized equation into harmonics, which would reduce it to a sequence of uncoupled linear ODEs in the spatial coordinate.
 
  • #6
Sorry for late reply!

Chestermiller said:
I'm not familiar with that analytic solution, but even if you know that, how would you then include oscillations.
The technique we use is to basically solve 5 linearized equations where velocity ##u## admits a potential ##\phi##. So we're inviscid, though we have viscous potential theory predictions as well. The 5 eqns are

$$
\nabla^2 \phi = 0 \,\,\, (\Omega)\\
\phi_n \cdot \hat n = 0 \,\,\,(\Sigma)\\
\int_\Gamma \phi_n = 0\,\,\,(\Gamma)\\
B[\phi_n] = 0 \,\,\, (\gamma)\\
-\phi_n''(s) - \cos^2\alpha \phi_n = \lambda^2 \phi \,\,\, (\Gamma)
$$
where ##\Omega## is liquid volume, ##\Sigma## is container walls, ##\Gamma## is equilibrium interface, ##\gamma## is equilibrium contact line, and ##\alpha## is equilibrium contact angle. Then those equations are continuity, no penetration, volume conservation, a contact line/angle condition defined by some operator ##B##, and a pressure balance among inertial pressure (linearized Bernoulli's) and Laplace pressure (linearized surface curvatures). A normal mode assumption in time is made, where the temporal component is ##\exp(i \lambda t)##.

Chestermiller said:
I would start out by writing the PDE for the behavior with oscillations, then linearize with respect to the oscillations, and then separate that into the non-linear equation you have already plus the linearized equation for the oscillations (based on the nonlinear static equation solution). Then I would resolve the linearized equation into harmonics, which would reduce it to a sequence of uncoupled linear ODEs in the spatial coordinate.

That sounds similar to what we've done, right? Clearly there is some difference since I end up with a single ODE rather than a system. Can you elaborate further or recommend literature here?
 
  • #7
joshmccraney said:
Sorry for late reply!The technique we use is to basically solve 5 linearized equations where velocity ##u## admits a potential ##\phi##. So we're inviscid, though we have viscous potential theory predictions as well. The 5 eqns are

$$
\nabla^2 \phi = 0 \,\,\, (\Omega)\\
\phi_n \cdot \hat n = 0 \,\,\,(\Sigma)\\
\int_\Gamma \phi_n = 0\,\,\,(\Gamma)\\
B[\phi_n] = 0 \,\,\, (\gamma)\\
-\phi_n''(s) - \cos^2\alpha \phi_n = \lambda^2 \phi \,\,\, (\Gamma)
$$
where ##\Omega## is liquid volume, ##\Sigma## is container walls, ##\Gamma## is equilibrium interface, ##\gamma## is equilibrium contact line, and ##\alpha## is equilibrium contact angle. Then those equations are continuity, no penetration, volume conservation, a contact line/angle condition defined by some operator ##B##, and a pressure balance among inertial pressure (linearized Bernoulli's) and Laplace pressure (linearized surface curvatures). A normal mode assumption in time is made, where the temporal component is ##\exp(i \lambda t)##.
That sounds similar to what we've done, right? Clearly there is some difference since I end up with a single ODE rather than a system. Can you elaborate further or recommend literature here?
Sorry, but I don't follow your analysis. Maybe someone else can help.
 

FAQ: Fluid Equilibrium Interface Shape

1. What is fluid equilibrium interface shape?

Fluid equilibrium interface shape refers to the shape that a fluid takes on when it is in a state of equilibrium, meaning that the forces acting on the fluid are balanced and there is no net movement of the fluid. This shape is determined by the surface tension and gravity acting on the fluid.

2. How is the shape of a fluid interface determined?

The shape of a fluid interface is determined by the balance of surface tension and gravity. Surface tension is the cohesive force between molecules at the surface of a liquid, while gravity pulls the fluid downward. The shape of the interface is determined by the relative strength of these forces.

3. What factors affect the equilibrium interface shape?

The equilibrium interface shape is affected by several factors, including the properties of the fluid (such as surface tension and density), the shape of the container holding the fluid, and the external forces acting on the fluid (such as gravity or external pressures).

4. How does surface tension affect the equilibrium interface shape?

Surface tension plays a crucial role in determining the equilibrium interface shape. A higher surface tension will result in a more curved interface, while a lower surface tension will result in a flatter interface. This is because surface tension acts to minimize the surface area of the fluid, causing it to take on a shape with the smallest surface area possible.

5. What is the significance of fluid equilibrium interface shape?

The fluid equilibrium interface shape has important implications in various fields, such as fluid mechanics, materials science, and biology. Understanding the shape of a fluid interface can help predict the behavior of fluids in different environments, and can also be used to design and optimize various systems and processes.

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