Fluid pressure at an interface

Or is u a vector and phi a matrix?I think I don't understand this...are you taking the inner product of a vector (u) and a scalar (phi)? Or is u a vector and phi a matrix?No, phi are basis functions (scalars), and u is a vector of unknownsNo, phi are basis functions (scalars), and u is a vector of unknownsOh. Then there isn't any reason that I can see that M should be symmetric.Oh. Then there isn't any reason that I can see that M should be symmetric.Gotcha. Thanks for all your help!Gotcha
  • #1
member 428835
Suppose we have an incompressible, viscous sessile drop subject to a time-dependent pressure field ##p## on a substrate. Let ##\mu## be dynamic viscosity, ##u## be the fluid velocity field, ##\kappa_{1/2}## curvatures of the fluid surface, ##\sigma## surface tension, ##\hat n## normals to the equilibrium surface, and ##\eta## the disturbed interface.

Disturbances to the equilibrium surface generate pressure gradients, and thereby flows. A pressure balance at the interfacial surface yields $$p-\mu \hat n \cdot(\nabla \otimes u) \cdot \hat n = - \sigma( \Delta_\Gamma \eta + (\kappa_1^2+\kappa_2^2)\eta)$$

The RHS is flow from the capillary pressure (Young-Laplace equation). The LHS is inertial pressure (first term) and viscous pressure (second term). I do not understand where the viscous pressure entered the pressure balance.

After googling I found this site: http://web.mit.edu/1.63/www/Lec-notes/Surfacetension/Lecture2.pdf

where equation (3) looks like the LHS, and if we look at their definition of ##T## we see there is a transpose velocity component (not shown in the pressure balance above, and the implication ##\hat n \cdot -p I \cdot \hat n = -p##)? Can someone help me understand this? Thanks!
 
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  • #2
I agree with the MIT analysis. What happened to the ##(\nabla \otimes u)^T## term in your equation for the stress tensor for a Newtonian fluid? Why is it missing?
 
  • #3
Chestermiller said:
I agree with the MIT analysis. What happened to the ##(\nabla \otimes u)^T## term in your equation for the stress tensor for a Newtonian fluid? Why is it missing?
It's from this paper, equation (4): https://pubs.rsc.org/en/content/articlepdf/2016/sm/c6sm01928e

Author does not mention it. Do you have an idea?
 
  • #5
Chestermiller said:
Sorry. I don't have access to that article...retired and all.
Attached is the Mathematical formulation. Everything before this is introduction, so not useful for my question.
 

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  • #6
Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.
 
  • #7
Chestermiller said:
Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.
So you think what they wrote is in fact wrong?
 
  • #8
Chestermiller said:
Isn't n dot (a tensor) dot n the same as n dot (the transpose of the tensor) dot n? If so, then all they are missing is a factor of 2.
Also, what would be a good way to know if ##\hat n \cdot \nabla u \cdot \hat n = \hat n \cdot (\nabla u)^T \cdot \hat n##? Like, where did your intuition come from?
 
  • #9
joshmccraney said:
Also, what would be a good way to know if ##\hat n \cdot \nabla u \cdot \hat n = \hat n \cdot (\nabla u)^T \cdot \hat n##? Like, where did your intuition come from?
Just evaluate it in Cartesian Coordinates and see whether it is correct.
 
  • #10
Chestermiller said:
Just evaluate it in Cartesian Coordinates and see whether it is correct.
Thinking about it and reviewing notes, perhaps we don't need to show it's true for cartesian. We know $$ \nabla u = D + \Omega : \\
D \equiv \frac{1}{2} (\nabla u + \nabla u ^T),
\Omega \equiv \frac{1}{2} (\nabla u - \nabla u ^T)$$
and we know ##D=D^T## and ##\Omega = -\Omega^T##. Then

$$n \cdot \nabla u \cdot n = n \cdot (D + \Omega) \cdot n$$

where we note $$n\cdot\Omega\cdot n = n_i \Omega_{ij} n_j = -n_i \Omega_{ji} n_j$$ so then ##-\Omega = \Omega \implies \Omega = 0.## Also we can show since ##D = D^T## that ##\nabla u ^T = \nabla u##. I think this proves what we seek to show, right?
 
  • #11
The transpose of the velocity gradient tensor is not equal to the velocity gradient tensor itself.
 
  • #12
Chestermiller said:
The transpose of the velocity gradient tensor is not equal to the velocity gradient tensor itself.
Can you elaborate please? I thought ##n \cdot \nabla u \cdot n = n \cdot ( \nabla u)^T \cdot n##.
 
  • #13
joshmccraney said:
Can you elaborate please? I thought ##n \cdot \nabla u \cdot n = n \cdot ( \nabla u)^T \cdot n##.
That is correct, but it doesn’t guarantee that the gradient of the velocity and its transpose are equal.
 
  • #14
Chestermiller said:
That is correct, but it doesn’t guarantee that the gradient of the velocity and its transpose are equal.
I must be missing something. Isn't it enough to show that ##(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n ##?
 
  • #15
You seem to be trying to prove that ##\nabla u=(\nabla u)^T##. Just because ##(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n ## is true doesn't necessarily mean that ##\nabla u=(\nabla u)^T##. As a matter of fact, the latter is not generally correct.
 
  • #16
Chestermiller said:
You seem to be trying to prove that ##\nabla u=(\nabla u)^T##. Just because ##(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n ## is true doesn't necessarily mean that ##\nabla u=(\nabla u)^T##. As a matter of fact, the latter is not generally correct.
Gotcha. But it is true that ##(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n ##, right?

Also, when I take the inner product $$M_{ij} = \int (n \cdot \nabla \nabla \phi_i \cdot n) \phi_j : u = \nabla \phi$$ it turns out ##M_{ij} \neq M_{ji}##. I have reason to think ##M## should be symmetric. What do you think?
 
  • #17
joshmccraney said:
Gotcha. But it is true that ##(n \cdot \nabla u \cdot n + n \cdot (\nabla u)^T \cdot n)/2 = n \cdot \nabla u \cdot n ##, right?
Yes. You can show this using Cartesian coordinates.
Also, when I take the inner product $$M_{ij} = \int (n \cdot \nabla \nabla \phi_i \cdot n) \phi_j : u = \nabla \phi$$ it turns out ##M_{ij} \neq M_{ji}##. I have reason to think ##M## should be symmetric. What do you think?
I think I don't understand this notation.
 
  • #18
Chestermiller said:
Yes. You can show this using Cartesian coordinates.
Yes, I did this for the problem I'm working on and it's true.

Chestermiller said:
I think I don't understand this notation.
Sorry, let me ask the question in a better way: is it true that the notation here are always equivalent ##\nabla \otimes \vec u = \nabla \vec u##?
 
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  • #19
joshmccraney said:
Yes, I did this for the problem I'm working on and it's true.Sorry, let me ask the question in a better way: is it true that the notation here are always equivalent ##\nabla \otimes \vec u = \nabla \vec u##?
That's my understanding
 
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  • #20
Thanks!
 

Related to Fluid pressure at an interface

1. What is fluid pressure at an interface?

Fluid pressure at an interface refers to the force per unit area exerted by a fluid on the surface of an object or another fluid at the boundary between the two. It is a result of the weight of the fluid above the interface and the force of gravity.

2. How is fluid pressure at an interface calculated?

Fluid pressure at an interface can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid at the interface.

3. What factors affect fluid pressure at an interface?

The factors that affect fluid pressure at an interface include the density of the fluid, the acceleration due to gravity, and the depth of the fluid at the interface. Additionally, the shape and size of the object or container containing the fluid can also impact the pressure at the interface.

4. How does fluid pressure at an interface change with depth?

Fluid pressure at an interface increases with depth as the weight of the fluid above the interface increases. This is due to the increased number of fluid molecules above the interface exerting a greater force on the surface.

5. What is the significance of fluid pressure at an interface?

Fluid pressure at an interface is important in various applications, such as in hydraulic systems, where it is used to control the movement of fluids. It also plays a crucial role in understanding the behavior of fluids in different environments, such as in oceans and in the atmosphere.

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