Normal stress boundary condition at fluid/vacuum interface

[Type1civ]

Homework Statement

Stuck on two similar problems:

"State the normal stress boundary condition at an interface
$$x_3-h(x_1,x_2,t)=0$$between an invisicid incompressible fluid and a vacuum. You may assume that the interface has a constant tension."

The second question in the same but the fluid is viscous.

2. The attempt at a solution
I worked out the normal to the surface to be:$$\mathbf{\hat n}=\frac{\partial h}{\partial t} -\mathbf{\hat x_3}+\frac{\partial h}{\partial x_1}\mathbf{\hat x_1}+\frac{\partial h}{\partial x_2}\mathbf{\hat x_2}$$so then the normal stress at the boundary much be equal and opposite from both media? But since one of the medium is a vacuum the stress will be zero. so:
$$\frac{\partial h}{\partial t} -\frac{\partial w}{\partial x_3}+\frac{\partial h}{\partial x_1}\frac{\partial u}{\partial x_1}+\frac{\partial h}{\partial x_2}\frac{\partial v}{\partial x_2}=0$$
where I am using the velocity of the fluid so that $\mathbf{u}=(u,v,w)$.

Not really sure how much of this is the correct line of thought, and also not sure how the viscosity should impact my result...

Thanks

 

[Chestermiller]

This is not the correct equation for a normal to the surface.

A differential position vector within the surface (i.e., tangent to the surface) is given by:

$$\vec{ds}=\vec{i_1}dx_1+\vec{i_2}dx_2+\left(\frac{∂h}{∂x_1}dx_1+\frac{∂h}{∂x_2}dx_2\right)\vec{i_3}=\left(\vec{i_1}+\frac{∂h}{∂x_1}\vec{i_3} \right)dx_1 +\left(\vec{i_2}+\frac{∂h}{∂x_2}\vec{i_3}\right)dx_2$$
Do you know how to use this relationship to get the equation for a unit normal to the surface? Hint: It involves taking a cross product.

Chet

 

[Type1civ]

Thanks for the reply,

The normal is just $$\mathbf{n}=\frac{\nabla F}{|\nabla F|}$$
isn't it? (where F is the function of the surface) so with my function (but no time dependence) my normal vector would be:
$$\mathbf{\hat n}=-\mathbf{\hat x_3}+\frac{\partial h}{\partial x_1}\mathbf{\hat x_1}+\frac{\partial h}{\partial x_2}\mathbf{\hat x_2}$$
where i haven't normalized it. Then I thought I could include the time dependence in the way I did in my original post, is this not correct?

 

[Chestermiller]

Thanks for the reply,

The normal is just $$\mathbf{n}=\frac{\nabla F}{|\nabla F|}$$
isn't it? (where F is the function of the surface) so with my function (but no time dependence) my normal vector would be:
$$\mathbf{\hat n}=-\mathbf{\hat x_3}+\frac{\partial h}{\partial x_1}\mathbf{\hat x_1}+\frac{\partial h}{\partial x_2}\mathbf{\hat x_2}$$
where i haven't normalized it. Then I thought I could include the time dependence in the way I did in my original post, is this not correct?

Oop sorry. You were right. But, the normal should not involve the time derivative. Besides, in your original post, the time derivative did not even multiply a unit vector.

You have to identify which side of the interface the fluid resides on. You are going to need to provide an expression for the unit normal drawn from the side of the interface that the fluid resides on to the side of the interface that the vacuum resides on. My guess that it is minus your normal (with a positive sign in front of the z unit vector). You also need to divide your normal by its magnitude to get the unit normal.

I think you should first consider the case where there is no tension in the membrane. Why? If you can't do that case, then you certainly won't be able to do the case with tension in the membrane.

According to the Cauchy stress relationship, the stress vector at the interface is equal to the stress tensor dotted with the unit normal. In the case of an inviscid fluid, the stress tensor is equal to the fluid pressure times the identity (metric) tensor. So, what does that give you for the stress vector, and, since that must match the zero pressure vacuum on the other side of the interface, what does that give you for the boundary condition.

Next, what is the dot product of the stress tensor and the unit normal for the more general case of a Newtonian fluid? First express this in terms of the σ's, and then substitute the velocity gradient terms.

To do the cases where the membrane has tension, there are different ways of approaching this. One involves taking a little curved surface "window" element with sides dx₁ and dx₂, and doing a force balance on this surface window. You may have learned how to do this a different way. Another approach?

Chet

 

[Type1civ]

Firstly, thank you so much for taking the time to help me, I need to get this done tonight and was worried that I had no chance of doing so.

According to the Cauchy stress relationship, the stress vector at the interface is equal to the stress tensor dotted with the unit normal. In the case of an inviscid fluid, the stress tensor is equal to the fluid pressure times the identity (metric) tensor. So, what does that give you for the stress vector, and, since that must match the zero pressure vacuum on the other side of the interface, what does that give you for the boundary condition.

Chet

Ok, so I would just end up with:
$$\mathbf{t}\cdot\mathbf{n}=(-1+\frac{\partial h}{\partial x_1}+\frac{\partial h}{\partial x_2})p=0$$
Since the components of stress tensor $\mathbf{t}$ are simply the pressure times the identity matrix.

To do the cases where the membrane has tension, there are different ways of approaching this. One involves taking a little curved surface "window" element with sides dx₁ and dx₂, and doing a force balance on this surface window. You may have learned how to do this a different way. Another approach?

I think for a viscous fluid I can simply add a a term which relates to the viscosity of the fluid so that my stress tensor is the same as before but with an extra an term relating to the shear stress added on: $$\mu\bigg(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\bigg)$$
Does this sound right?

 

[Type1civ]

Sorry I have mixed notation here I should have written my normal dotted with my stress tensor as
$$t_{ij}n_j=0$$

 

[Chestermiller]

You showed correctly that, if there is no tension in the membrane, the fluid pressure at the boundary is zero for an inviscid fluid.

Now, to be more precise, the unit normal drawn from the fluid side of the membrane to the vacuum side of the membrane is:
$$\vec{n}_{fv}=\frac{\vec{i}_3-(∂h/∂x_1)\vec{i}_1-(∂h/∂x_2)\vec{i}_2}{\sqrt{1+(∂h/∂x_1)^2+(∂h/∂x_2)^2}}$$
Conversely, the unit normal drawn from the vacuum side of the membrane to the fluid side of the membrane is:
$$\vec{n}_{vf}=-\frac{\vec{i}_3-(∂h/∂x_1)\vec{i}_1-(∂h/∂x_2)\vec{i}_2}{\sqrt{1+(∂h/∂x_1)^2+(∂h/∂x_2)^2}}$$

The stress tensor on the fluid side of the membrane is given by:
$$\vec{t}_f=-p(\vec{i}_1\vec{i}_1+\vec{i}_2\vec{i}_2+\vec{i}_3\vec{i}_3)+μ\left( \frac{∂u_k}{∂x_j}+\frac{∂u_j}{∂x_k}\right)\vec{i}_k\vec{i}_j$$
where Einstein summation is implied in the second term.

The stress tensor on the vacuum side of the membrane is given by:
$$\vec{t}_v=\vec{0}$$
The stress vector exerted by an inviscid fluid upon the fluid side of the membrane is given by $$\vec{t}_f\centerdot \vec{n}_{vf}=p\vec{n}_{fv}$$
This must equal the stress vector exerted by the vacuum on the vacuum side of the membrane which is given by $$\vec{t}_v\centerdot \vec{n}_{fv}=\vec{0}$$
Therefore, p = 0 at the boundary for an inviscid fluid (no membrane stress).

For a viscous Newtonian fluid, the normal component of the stress vector exerted by the fluid upon the fluid side of the membrane is given by:
$\vec{n}_{vf} \centerdot \vec{t}_f\centerdot \vec{n}_{vf}=-p+2μ\frac{\left(\frac{∂h}{∂x_1}\right)^2\frac{∂u_1}{∂x_1}+ \left(\frac{∂h}{∂x_2}\right)^2\frac{∂u_2}{∂x_2}+\frac{∂u_3}{∂x_3}+\left(\frac{∂h}{∂x_1}\right)\left(\frac{∂h}{∂x_2}\right)\left( \frac{∂u_1}{∂x_2}+\frac{∂u_2}{∂x_1}\right) -\left(\frac{∂h}{∂x_1}\right)\left( \frac{∂u_1}{∂x_3}+\frac{∂u_3}{∂x_1}\right) -\left(\frac{∂h}{∂x_2}\right)\left( \frac{∂u_2}{∂x_3}+\frac{∂u_3}{∂x_2}\right)}{1+(∂h/∂x_1)^2+(∂h/∂x_2)^2}$

Since the normal component of stress on the vacuum side is zero, we must have that:
$$p=2μ\frac{\left(\frac{∂h}{∂x_1}\right)^2\frac{∂u_1}{∂x_1}+ \left(\frac{∂h}{∂x_2}\right)^2\frac{∂u_2}{∂x_2}+\frac{∂u_3}{∂x_3}+\left(\frac{∂h}{∂x_1}\right)\left(\frac{∂h}{∂x_2}\right)\left( \frac{∂u_1}{∂x_2}+\frac{∂u_2}{∂x_1}\right) -\left(\frac{∂h}{∂x_1}\right)\left( \frac{∂u_1}{∂x_3}+\frac{∂u_3}{∂x_1}\right) -\left(\frac{∂h}{∂x_2}\right)\left( \frac{∂u_2}{∂x_3}+\frac{∂u_3}{∂x_2}\right)}{1+(∂h/∂x_1)^2+(∂h/∂x_2)^2}$$
This is the boundary condition if the membrane has no stress in it.

Doing the problem with membrane stress included is a little more challanging, and I'm going to stop here for now to let you digest what I've written. Any thoughts on how to include the membrane stress.

Chet