# I When to use which dimensionless number

1. Dec 5, 2017

### joshmccraney

Yes, you're right. This is a canonical problem in capillary driven corner flows. There are several papers describing what I posted, though perhaps slightly more dense that what I wrote? And the NASA flight experiment you refer to is the topic of my grant proposal due Dec 15. In fact, my previous adviser at a different university than that I attend now was the PI for the flight experiment you refer to: CFE ICF flight experiments.

Hahahahhaa small world. So, as you may wonder, how did this question initially surface about dimensionless numbers? My current adviser asked me for Reynolds numbers and Ohnesorge numbers for the very flight experiments you reference (CFE ICF). I was confused because I've never considered Reynolds numbers for that flow since the scaling didn't automatically provide any. I was curious what people on these forums thought

2. Dec 5, 2017

### Staff: Mentor

I'm sorry Josh. I can't make sense out of this figure. I'm more confused than ever. Can you please try again?

3. Dec 5, 2017

### joshmccraney

Here's a sketch I drew a while ago. I probably should've used this earlier. The corner angle measures $2\alpha$ (not shown).

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4. Dec 5, 2017

### Staff: Mentor

This is much better. As I understand it, you have a trough with liquid flowing down the trough. It is a transient situation, with an advancing wetting front. You are introducing fluid at the left. Does this capture it?

5. Dec 5, 2017

### joshmccraney

Your understanding is correct! Yes, it's transient, and some of the flows have an advancing front, and others are pinned. I think the scales work out; what do you think? I suppose you could specify a flow-rate condition at a specified height at an end, but I don't think that would change the analysis would it?

6. Dec 5, 2017

7. Dec 5, 2017

### Staff: Mentor

Is the flow supposed to be a uniform velocity profile at the inlet? Is that what is specified? I assume this is set up in cylindrical coordinates, correct? My first step would be to parameterize the free surface shape, and get surface tangents, unit normal, and curvature.

8. Dec 5, 2017

### joshmccraney

I've seen all this in papers, but sometimes you do things differently. If you want to see how others have done it I am happy to send you the publication but I wouldn't mind reworking this with you. Again, you've done things differently in the past and I always learn something.

9. Dec 6, 2017

### Andy Resnick

I'm familiar with these experiments :)

Yeah, it's tricky because there's a lot going on. For the CFE-ICF experiments concerned with two-phase bubbly flow, I would expect the capillary number to be most important, and the Oh and We numbers to be less relevant. To be sure, two-phase flow regimes are segregated using terms involving the Reynolds number:

http://cdn.intechopen.com/pdfs/40637/InTech-Two_phase_flow.pdf

If the liquid phase does not completely wet the solid, then the real issue is contact line motion- I don't think that problem can be scaled to generate a dimensionless number (similarity parameter).

https://www.cambridge.org/core/jour...erior-corner/8C88D0772C50153457B155B8B1DDB029

10. Dec 6, 2017

### joshmccraney

That's a good link for dimensionless numbers.

Oh yes, when I was Mark's (Weislogel) student he gave me his thesis to familiarize myself with the subject, the bulk of which is in this publication. I've read this paper once or twice ;)

11. Dec 6, 2017

### Staff: Mentor

OK Josh. In this spirit, this is how I would obtain the Young-Laplace equation for the free surface:
Using cylindrical coordinates, let $R(t, \theta, z)$ represent the instantaneous radial coordinate of the free surface as a function of t, $\theta$, and z. This fully describes the shape of the free surface at any time t. Let $\mathbf{r}$ represent a position vector drawn from the origin (lower left hand corner of trough) to an arbitrary point $[R(t,\theta,z),\ \theta,\ z]$ on the free surface at time t. We have:$$\mathbf{r}=R(\theta, z)\hat{r}+z\hat{z}\tag{1}$$where, for convenience, we have suppressed the dependence on t. From Eqn. 1, an arbitrary differential position vector within the "plane" of the free surface is then given by:
$$\mathbf{dr}=\left(\frac{\partial R}{\partial \theta}d\theta+\frac{\partial R}{\partial z}dz\right)\hat{r}+R\hat{\theta}d\theta+\hat{z}dz\tag{2}$$This equation can be rearranged to give:
$$\mathbf{dr}=\hat{a}_{\theta}d\theta+\hat{a}_zdz\tag{3}$$with $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$where $\hat{a}_{\theta}$ and $\hat{a}_z$ are "coordinate basis vectors" for the free surface, and are tangent to the surface (i.e., lie within the "plane" of the surface) at all locations.

I think I'll stop here for now and give you a chance to digest these equations, particularly Eqns. 3, which are very fundamental.

I still don't think I fully understand the problem that you are trying to solve, and I don't see where there are any natural length scales for the problem, unless the end at z = L is a natural barrier. However, for now, I'm content to focus on the free surface. Is the angle of the trough usually taken to be pretty small?

12. Dec 9, 2017

### joshmccraney

I agree, Young-Laplace seems correct. I say this because surface tension dominates. However, in other flows, say the canonical flow down an inclined plane, we don't use Young-Laplace. So when do you know to use this equation?

Ok, I understand why you've selected this coordinate system, how to compute $\mathbf{dr}$, and that it lies along the plane of the surface. Could you elaborate on the basis vectors, specifically how you selected them (or perhaps how the equation suggests them)?

That's a good thing you don't understand the problem, because so far we haven't prescribed any boundary conditions, so as is we're just analyzing a general flow, right? But, what if we had draining at $z=\pm L\implies \mathbf{r}=0$ there. Then we have a characteristic length scale and height scale, right? And yea, typically $R\ll L$ and $\alpha \sim 30^\circ$. But you can make whatever assumptions you want, I'm just here to learn from you.

13. Dec 9, 2017

### Staff: Mentor

Whenever the free surface has significant curvature, Young-Laplace needs to be used.
This all comes from Gauss' analysis of surfaces (2D manifolds). He realized that, if you have two arbitrary coordinates within a surface in 3D space (describing lines of, say, constant x1 and x2) you could represent differential position vectors joining closely neighboring points within the surface by an equation of the general form $d\vec{r}=\vec{a_{1}}(x_1,x_2)dx_1+\vec{a_2}(x_1,x_2)dx_2$ (with the a1 vector pointing locally along the lines of constant x2, and the a2 vectors pointing locally along the lines of constant x1). You could then use this to establish all the metrical properties of the 2D surface. You have basically set up a curvilinear coordinate system within the surface. The basis vectors automatically suggest themselves once you specify the spatial coordinates used to grid-up the surface.

So, are you ready to move to the next step now?

14. Dec 9, 2017

### joshmccraney

Makes a lot of sense! Yes, I'm ready to see what's next!

15. Dec 9, 2017

### Staff: Mentor

What we are going to do next is focus on a tiny window of curved free surface between the grid lines $\theta$ and $\theta + d\theta$ in the $\theta$ in-plane direction, and the grid lines z and z + dz in the z in-plane direction. The opposite edges of this tiny window are not quite parallel to one another, nor are they quite perpendicular to their adjacent edges. We are going to do a force balance on the tiny window to derive the relationship between the normal force exerted by the pressure differential across the window and the surface tension forces at the edges of the window.

To do this force balance, we are going to need to derive equations for the two unit normal in-plane vectors to the grid lines of constant $\theta$ and z within the surface. We need these because the surface tension forces at the edges of the window act perpendicular to the edges.

To implement this derivation, we first define two in-plane "reciprocal coordinate basis vectors" $\hat{a}^{\theta}$ and $\hat{a}^z$ within the surface as follows:

$\hat{a}^{\theta}\centerdot \hat{a}_z=0$ and $\hat{a}^{\theta}\centerdot \hat{a}_{\theta}=1$
and
$\hat{a}^{z}\centerdot \hat{a}_{\theta}=0$ and $\hat{a}^{z}\centerdot \hat{a}_{z}=1$

So $\hat{a}^{\theta}$ is perpendicular to the lines of constant z, and its orientation and magnitude are such that it is at a typically small angle to the lines of constant $\theta$, and the product of its magnitude with that of $\hat{a}_{\theta}$ times the cosine of the angle between them is equal to unity. A similar geometric interpretation applies to $\hat{a}^z$.

In this derivation, what we are going to do is express $\hat{a}^{\theta}$ and $\hat{a}^z$ each as a linear combination of $\hat{a}_{\theta}$ and $\hat{a}_z$. For $\hat{a}^{\theta}$, we first write:$$\hat{a}^{\theta}=A\hat{a}_{\theta}+B\hat{a}_z$$where A and B are coefficients (to be determined). If we dot $\hat{a}^{\theta}$ with $\hat{a}_z$ and $\hat{a}_{\theta}$, we obtain:
$$\hat{a}^{\theta}\centerdot \hat{a_z}=g_{\theta z}A+g_{zz}B=0\tag{4a}$$and$$\hat{a}^{\theta}\centerdot \hat{a}_{\theta}=g_{\theta \theta}A+g_{\theta z}B=1\tag{4b}$$where the three "metrical coefficients" for the surface, $g_{\theta \theta}$, $g_{\theta z}$, and $g_{zz}$, are defined by:
$$g_{\theta \theta}=\hat{a}_{\theta}\centerdot \hat{a}_{\theta}\tag{5a}$$
$$g_{\theta z}=\hat{a}_{\theta}\centerdot \hat{a}_{z}\tag{5b}$$and $$g_{zz}=\hat{a}_{z}\centerdot \hat{a}_{z}\tag{5c}$$
Now its your turn. Please solve Eqns. 4 for the coefficients A and B in terms of $g_{\theta \theta}$, $g_{\theta z}$, and $g_{zz}$, and then write out the equation for $\hat{a}^{\theta}$ in terms of these.

16. Dec 15, 2017

### joshmccraney

Ahhh this is related to covariant and contravariant vectors, right? I think I've done this before but it was about a year ago. At any rate, we have $$A = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\\ B = \frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\implies\\ \hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z.$$

Sorry my response took so long. I should now be able to respond much faster (very busy week for me).

17. Dec 15, 2017

### Staff: Mentor

Good. Next, determine the magnitude of $\hat{a}^{\theta}$ in terms of the g's. Then, divide $\hat{a}^{\theta}$ by its magnitude to get the in-plane unit vector $\hat{u}^{\theta}$ in the direction perpendicular to $\hat{a}^z$.

18. Dec 15, 2017

### joshmccraney

$$\hat{u}^\theta = \frac{\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z}{\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}}\\ = \left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \right)\sqrt{\frac{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}{g_{zz}^2+g_{\theta z}^2}}\\ =\frac{g_{zz}\hat{a}_\theta+g_{\theta z}\hat{a}_z}{\sqrt{g_{zz}^2+g_{\theta z}^2}}$$
though I loosely let the absolute values cancel the denominators in the last step, so perhaps one term is off by a sign?

19. Dec 15, 2017

### Staff: Mentor

That's not what I get. What do you get for the magnitude?

20. Dec 15, 2017

### joshmccraney

$$\sqrt{\frac{g_{zz}^2+g_{\theta z}^2}{(g_{zz}g_{\theta\theta}-g_{\theta z}^2)^2}}$$