When to use which dimensionless number

[member 428835]

I confirm all those components.

Ok, then here is what follows:

Notice I used the FullSimplify command, so ideally this is concise. It's not the same as yours (I input that in already). My results based on post 84

 

[Chestermiller]

Ok, then here is what follows:View attachment 220354
Notice I used the FullSimplify command, so ideally this is concise. It's not the same as yours (I input that in already). My results based on post 84

As best I can tell, this is equivalent to what I had.

 

[member 428835]

As best I can tell, this is equivalent to what I had.

Shoot, maybe I typed it into Mathematica wrong. So, if both are correct, then we're good?

 

[Chestermiller]

Shoot, maybe I typed it into Mathematica wrong. So, if both are correct, then we're good?

Yes. But, I'd also like to include the viscous part of the stress in the boundary condition at the free surface. In terms of the components of the stress tensor (in cylindrical coordinates) and our equation for the normal to the free surface, what is the traction vector at the free surface?

 

[member 428835]

Yes. But, I'd also like to include the viscous part of the stress in the boundary condition at the free surface. In terms of the components of the stress tensor (in cylindrical coordinates) and our equation for the normal to the free surface, what is the traction vector at the free surface?

If the stress tensor is $\sigma$ then are you asking for $\sigma \cdot \hat N$? Shouldn't we instead take $\sigma \cdot \hat u_z$ and $\sigma \cdot \hat u_\theta$ since we're concerned with finding shear? Evidently the stress tensor $\sigma$ has components

where we can obviously omit the $\nabla\cdot\hat V$ terms (incompressible). So the matrix appears as
$$
\sigma =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
$$
and $\hat u_z,\hat u_\theta$ have bee previously defined. What do you think?

 

[Chestermiller]

If the stress tensor is $\sigma$ then are you asking for $\sigma \cdot \hat N$? Shouldn't we instead take $\sigma \cdot \hat u_z$ and $\sigma \cdot \hat u_\theta$ since we're concerned with finding shear? Evidently the stress tensor $\sigma$ has components View attachment 220423
where we can obviously omit the $\nabla\cdot\hat V$ terms (incompressible). So the matrix appears as
$$
\sigma =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
$$
and $\hat u_z,\hat u_\theta$ have bee previously defined. What do you think?

I'd like to see (in terms of the tau's),$$\hat{\sigma}\centerdot \hat{n}$$The above is the viscous stress (traction) vector at the interface
$$(\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n}$$The above is the normal viscous stress component at the interface
$$\hat{\sigma}\centerdot \hat{n}-((\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n})\hat{n}$$The above is the viscous shear stress at the interface (which must be zero)

 

[member 428835]

I'd like to see (in terms of the tau's),$$\hat{\sigma}\centerdot \hat{n}$$The above is the viscous stress (traction) vector at the interface
$$(\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n}$$The above is the normal viscous stress component at the interface
$$\hat{\sigma}\centerdot \hat{n}-((\hat{\sigma}\centerdot \hat{n})\centerdot \hat{n})\hat{n}$$The above is the viscous shear stress at the interface (which must be zero)

Gotcha. So
$$
\sigma \cdot \hat N =
\begin{bmatrix}
\tau_{rr} & \tau_{r\theta} & \tau_{rz}\\
\tau_{r\theta} & \tau_{\theta\theta} & \tau_{\theta z}\\
\tau_{rz} & \tau_{\theta z} & \tau_{zz}
\end{bmatrix}
\cdot
\begin{bmatrix}
R\\
-R_\theta\\
-R_zR
\end{bmatrix}
$$
where below is $\sigma \cdot \hat N$

and below here is $((\sigma\cdot\hat N)\cdot\hat N)\hat N$

 

[member 428835]

How does this look Chet?

 

[Chestermiller]

I don't understand the things that are bolded and the things that are not bolded. I get the following for the stress vector:

$$\hat{\sigma}\centerdot \hat{n}=(\sigma_{rr}n_r+\sigma_{r\theta}n_{\theta}+\sigma_{rz}n_z)\hat{r}+(\sigma_{r\theta}n_r+\sigma_{\theta \theta}n_{\theta}+\sigma_{\theta z}n_z)\hat{\theta}+(\sigma_{rz}n_r+\sigma_{\theta z}n_{\theta}+\sigma_{zz}n_z)\hat{z}$$
And, for the normal component of fluid stress at the interface, I get:
$$\hat{n}\centerdot \hat{\sigma}\centerdot \hat{n}=\sigma_{rr}(n_r)^2+\sigma_{\theta \theta}(n_{\theta})^2+\sigma_{zz}(n_z)^2+2\sigma_{r\theta}n_rn_\theta+2\sigma_{rz}n_rn_z+2\sigma_{\theta z}n_{\theta}n_z$$

 

[member 428835]

I don't understand the things that are bolded and the things that are not bolded.

Yea sorry, it's just how Mathematica outputs the symbols, but there is no significance with bold and unbold letters.

I get the following for the stress vector:

$$\hat{\sigma}\centerdot \hat{n}=(\sigma_{rr}n_r+\sigma_{r\theta}n_{\theta}+\sigma_{rz}n_z)\hat{r}+(\sigma_{r\theta}n_r+\sigma_{\theta \theta}n_{\theta}+\sigma_{\theta z}n_z)\hat{\theta}+(\sigma_{rz}n_r+\sigma_{\theta z}n_{\theta}+\sigma_{zz}n_z)\hat{z}$$

Awesome, that's what I get too (sorry, in the future I'll write everything in PF rather than snippets of Mathematica). It seems from post 97 that what I have as $\tau_{ij}$ you have as $\sigma_{ij}$ and you write $\hat n = \langle n_r, n_\theta, n_z\rangle$ where I wrote $\hat n = \langle R, -R_\theta, -R R_z \rangle$. So we get the same results.

And, for the normal component of fluid stress at the interface, I get:
$$\hat{n}\centerdot \hat{\sigma}\centerdot \hat{n}=\sigma_{rr}(n_r)^2+\sigma_{\theta \theta}(n_{\theta})^2+\sigma_{zz}(n_z)^2+2\sigma_{r\theta}n_rn_\theta+2\sigma_{rz}n_rn_z+2\sigma_{\theta z}n_{\theta}n_z$$

Ok, I'm getting $$
R^2(\sigma_{rr}-2\sigma_{rz}R_z+\sigma_{zz}R_z^2)-2R(\sigma_{r\theta}-\sigma_{\theta z}R_z)R_\theta+\sigma_{\theta\theta}R_\theta^2
$$
where I've adopted your notation using $\sigma_{ij}$ instead of $\tau_{ij}$, where the normal vector is written explicitly. I think we're getting the same thing here.

 

[member 428835]

What's next Chet? Hopefully you're not too bored of this yet.

 

[Chestermiller]

What's next Chet? Hopefully you're not too bored of this yet.

OK. I just wanted to make sure we evaluated the normal stress jump across the free surface, including both the pressure and the viscous contribution, and I wanted to show how to get the shear stress components at the free surface so that we could set them equal to zero in the fluid mechanics boundary conditions.

Now, I think we're done with the stress boundary conditions at the free surface, and we're back to trying to figure out how to reduce the equations of continuity and the Navier Stokes equations to dimensionless for for this problem. The game plan is to let the differential equations suggest to us the dimensionless groups to use. It seems like we've gravitated toward using cylindrical coordinates, although we could also do the stress boundary conditions (i.e., what we just did) at the free surface in spherical coordinates. Please write out the equation of continuity and the NS equations for whichever coordinate system you prefer, and with whatever terms you judge can be eliminated for whatever reason.

Chet

 

[member 428835]

Please write out the equation of continuity and the NS equations for whichever coordinate system you prefer, and with whatever terms you judge can be eliminated for whatever reason.
Chet

Okay, first I'll start by scaling characteristic lengths, velocities, and time. Assuming the wedge length is $L$ we know $z\sim L$. Conservation of mass in integral form reduces to $\partial_t A = -\partial_z(Aw)$, where $A$ is the cross section of the wedge. I can show details but I think this is relatively direct. This implies $t\sim L/w$. I think we can get scaling for $w$ by working with NS.

NS in general vector form is

$$\rho\frac{D \vec u}{Dt} = -\nabla P + \mu\nabla^2\vec u + \rho \vec g.$$

First assume velocity is small, so that non-linear velocity terms vanish. This simplifies the above to

$$\rho\frac{\partial \vec u}{\partial t} = -\nabla P + \mu\nabla^2\vec u + \rho \vec g.$$

Next assume pressure scales according to Young/Laplace equation $P\sim\sigma/R^2$ where $R$ is a characteristic radius and assume gravity scales as $g$. Assume $g<<\sigma/(R^2\rho)$ (small Bond number). Then NS further reduces to

$$\frac{\partial \vec u}{\partial t} = -\frac{1}{\rho}\nabla P + \nu\nabla^2\vec u.$$

My next approach would be to assume that flow in the $z$ direction dominates over radial and angular velocities, so the vector equation reduces at first order to

$$\frac{\partial w}{\partial t} = -\frac{1}{\rho}\partial_zP + \nu\nabla^2 w.$$

If we assume quasi-steady flow, then the above reduces to

$$\partial_zP = \mu\nabla^2 w.$$

The Laplacian can be divided into three scaled parts: $1/R^2,1/(R\theta)^2,1/L^2$. I would assume a lubrication approximation: $L>R\implies 1/L^2\ll 1/R^2$. I'm unsure how to work with $\theta$ other than saying $\theta\sim\alpha$. Then the Laplacian terms to leading order scale as $1/R^2,1/(R\alpha)^2$. Which one is dominant would require some physical reasoning. Assuming $\nabla^2\sim 1/R^2$ implies

$$\frac{\sigma}{R^2L} \sim \mu\frac{w}{R^2}\implies\\
w\sim \frac{\sigma}{L\mu} .$$

How does this look so far?

 

[Chestermiller]

Okay, first I'll start by scaling characteristic lengths, velocities, and time. Assuming the wedge length is $L$ we know $z\sim L$. Conservation of mass in integral form reduces to $\partial_t A = -\partial_z(Aw)$, where $A$ is the cross section of the wedge. I can show details but I think this is relatively direct. This implies $t\sim L/w$. I think we can get scaling for $w$ by working with NS.

NS in general vector form is

$$\rho\frac{D \vec u}{Dt} = -\nabla P + \mu\nabla^2\vec u + \rho \vec g.$$

First assume velocity is small, so that non-linear velocity terms vanish. This simplifies the above to

$$\rho\frac{\partial \vec u}{\partial t} = -\nabla P + \mu\nabla^2\vec u + \rho \vec g.$$

Next assume pressure scales according to Young/Laplace equation $P\sim\sigma/R^2$ where $R$ is a characteristic radius and assume gravity scales as $g$. Assume $g<<\sigma/(R^2\rho)$ (small Bond number). Then NS further reduces to

$$\frac{\partial \vec u}{\partial t} = -\frac{1}{\rho}\nabla P + \nu\nabla^2\vec u.$$

My next approach would be to assume that flow in the $z$ direction dominates over radial and angular velocities, so the vector equation reduces at first order to

$$\frac{\partial w}{\partial t} = -\frac{1}{\rho}\partial_zP + \nu\nabla^2 w.$$

If we assume quasi-steady flow, then the above reduces to

$$\partial_zP = \mu\nabla^2 w.$$

The Laplacian can be divided into three scaled parts: $1/R^2,1/(R\theta)^2,1/L^2$. I would assume a lubrication approximation: $L>R\implies 1/L^2\ll 1/R^2$. I'm unsure how to work with $\theta$ other than saying $\theta\sim\alpha$. Then the Laplacian terms to leading order scale as $1/R^2,1/(R\alpha)^2$. Which one is dominant would require some physical reasoning. Assuming $\nabla^2\sim 1/R^2$ implies

$$\frac{\sigma}{R^2L} \sim \mu\frac{w}{R^2}\implies\\
w\sim \frac{\sigma}{L\mu} .$$

How does this look so far?

Yikes. I'm not so sure I'm comfortable with all the approximations you made. Here are my comments

1. I realize the flow is in a wedge, but I really don't get what you did with the continuity equation.
2. Since the angle of the wedge is fixed, wouldn't it make more sense to use cylindrical coordinates?
3. Since the volume of liquid is constant, wouldn't it make more sense that the characteristic length would be the radius of a sphere formed by the fluid.
4. I don't understand your simplifications of the equation of motion. You seem to have gone to creeping flow in the z direction (presumably the vertical direction). And you seem to be neglecting the components in the other directions.
5. Before we get to the reduction of the equations to dimensionless form, I would first like to agree on the dimensional equations (including boundary conditions). I will then show you the powerful cookbook methodology I learned at Michigan for reducing the equations to dimensionless form.

 

[member 428835]

Yikes. I'm not so sure I'm comfortable with all the approximations you made. Here are my comments

1. I realize the flow is in a wedge, but I really don't get what you did with the continuity equation.
2. Since the angle of the wedge is fixed, wouldn't it make more sense to use cylindrical coordinates?
3. Since the volume of liquid is constant, wouldn't it make more sense that the characteristic length would be the radius of a sphere formed by the fluid.
4. I don't understand your simplifications of the equation of motion. You seem to have gone to creeping flow in the z direction (presumably the vertical direction). And you seem to be neglecting the components in the other directions.
5. Before we get to the reduction of the equations to dimensionless form, I would first like to agree on the dimensional equations (including boundary conditions). I will then show you the powerful cookbook methodology I learned at Michigan for reducing the equations to dimensionless form.

1. I didn't use continuity in differential form (at least not $\nabla \cdot \vec u = 0$). I used a control volume approach. Should I not have?
2. I was using cylindrical coordinates (notice my arguments for the Laplacian). But I ended up using just the $z$ direction.
3. Is the volume constant? I suppose we could consider a spreading droplet of water. I was imagining a set up where we could drain fluid at some downstream point, but perhaps we can talk more on this later?
4. Yea, I definitely was assuming creeping flow, and neglected flow in $r$ and $\theta$.
5. I'd like to learn this! Ok, so what's next? Shall I give it another go or would you like to take over?

 

[Chestermiller]

1. I didn't use continuity in differential form (at least not $\nabla \cdot \vec u = 0$). I used a control volume approach. Should I not have?
2. I was using cylindrical coordinates (notice my arguments for the Laplacian). But I ended up using just the $z$ direction.
3. Is the volume constant? I suppose we could consider a spreading droplet of water. I was imagining a set up where we could drain fluid at some downstream point, but perhaps we can talk more on this later?
4. Yea, I definitely was assuming creeping flow, and neglected flow in $r$ and $\theta$.
5. I'd like to learn this! Ok, so what's next? Shall I give it another go or would you like to take over?

So you are saying you have a trough with a very viscous fluid flowing down the trough? And no new fluid is being introduced on the high end?

 

[member 428835]

So you are saying you have a trough with a very viscous fluid flowing down the trough? And no new fluid is being introduced on the high end?

I was assuming the flow was very viscous. Did I enforce that no new fluid is being introduced? If so, this was not my intent (although we could do that too).

So in your analysis, how would you do things?

 

[Chestermiller]

I was assuming the flow was very viscous. Did I enforce that no new fluid is being introduced? If so, this was not my intent (although we could do that too).

So in your analysis, how would you do things?

Have you solved this without surface tension yet? If not, what do you think that solution looks like?

 

[member 428835]

Have you solved this without surface tension yet? If not, what do you think that solution looks like?

By "this" are you referring to the post in 103? If so, well the governing equation looks identical to that of Poiseuille flow, but obviously with a different geometry and BC, and the pressure I don't think would be constant here, though without surface tension then yes.

The way I understand the problem, flow is either driven by capillary forces or gravity (or perhaps both). The significance of each would depend on scaling. What do you think?

 

[Chestermiller]

By "this" are you referring to the post in 103? If so, well the governing equation looks identical to that of Poiseuille flow, but obviously with a different geometry and BC, and the pressure I don't think would be constant here, though without surface tension then yes.

The way I understand the problem, flow is either driven by capillary forces or gravity (or perhaps both). The significance of each would depend on scaling. What do you think?

By "this," I meant the overall problem. I have trouble seeing how capillary forces are going to be important unless the wedge angle is very small (and the height of water in the trough is small) or, at the very leading edge of the fluid as it advances in the trough. I'm going to try to set up this problem without surface tension an see where it takes me. I'll assume that the height of fluid in the trough is changing very gradually with axial position along the trough. Is that the kind of approximation you have been using?

 

[member 428835]

By "this," I meant the overall problem. I have trouble seeing how capillary forces are going to be important unless the wedge angle is very small (and the height of water in the trough is small) or, at the very leading edge of the fluid as it advances in the trough. I'm going to try to set up this problem without surface tension an see where it takes me. I'll assume that the height of fluid in the trough is changing very gradually with axial position along the trough. Is that the kind of approximation you have been using?

I think this thread was started considering capillary-driven flows, where gravity is ignored. There's no induced pressure gradient from external forces and no gravity, so isn't a capillary pressure force the only mechanism for driving flow?

This being said, I like the idea of including inertia! Since gravity is absent, surface tension (and capillarity) is relevant, right? I definitely used length much larger than height. If it's okay with you, can we proceed with surface tension?

 

[Chestermiller]

I think this thread was started considering capillary-driven flows, where gravity is ignored. There's no induced pressure gradient from external forces and no gravity, so isn't a capillary pressure force the only mechanism for driving flow?

This being said, I like the idea of including inertia! Since gravity is absent, surface tension (and capillarity) is relevant, right? I definitely used length much larger than height. If it's okay with you, can we proceed with surface tension?

Sorry Josh. I'm totally confused by this. Apparently, I just don't understand the essence of the problem.

 

[member 428835]

Sorry Josh. I'm totally confused by this. Apparently, I just don't understand the essence of the problem.

We can continue with your interpretation since I enjoy learning from you.

What I was imagining is a low-gravity situation, where the pressure gradient drives flow in a wedge but the pressure gradient is interpreted in the Young-Laplace equation since surface tension is relevant.

 

[member 428835]

Were you still interested in doing this Chet?

 

[Chestermiller]

Were you still interested in doing this Chet?

My energy for doing this has waned. But, here is the game plan I would use.

Assume that the liquid is very viscous and flowing very slowly. Assume that the surface of the liquid is horizontal in the direction perpendicular to the trough axis, so the surface elevation h is a function only of z. Use this to determine the local pressure gradient in the z direction. The shear stress on the fluid flowing down the trough is zero at the free surface, and the cross section of fluid at each axial location is basically half a rhombus. So the local pressure gradient would be the same as if you had a full rhombic closed channel with twice the volumetric flow rate. So, look up the relationship for fully developed laminar flow in a duct of rhombic cross section. Do a differential mass balance on the section of trough between z and z plus delta z to determine the rate of change of the liquid surface height with time as a function of z.