When to use which dimensionless number

[Chestermiller]

I'm with you, and agree with everything you've said so far!

OK. Please write down the force balance on the window, and divide by dtheta and dz. What do you get?

 

[member 428835]

OK. Please write down the force balance on the window, and divide by dtheta and dz. What do you get?

$$
\left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z+
\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}-\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}+
p\hat{a}_{\theta}\times \hat{a}_z d\theta dz = \vec 0\\

\frac{\partial}{\partial z}\left(\sigma \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\frac{\partial}{\partial \theta}\left(\sigma \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{a}_{\theta}\times \hat{a}_z = \vec 0
$$

where the rhs is 0 since the membrane does not have any mass. Something doesn't seem right though: the pressure term is the only term that is normal to the surface. If my work above is correct, then there would not be anything to balance it, implying $p=0$. What do you think?

 

[Chestermiller]

$$
\left.\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_{z+dz} \left.-\sigma d\theta\, \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right|_z+
\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta+d\theta}-\left.\sigma dz\, \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right|_{\theta}+
p\hat{a}_{\theta}\times \hat{a}_z d\theta dz = \vec 0\\

\frac{\partial}{\partial z}\left(\sigma \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\frac{\partial}{\partial \theta}\left(\sigma \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{a}_{\theta}\times \hat{a}_z = \vec 0
$$

where the rhs is 0 since the membrane does not have any mass. Something doesn't seem right though: the pressure term is the only term that is normal to the surface. If my work above is correct, then there would not be anything to balance it, implying $p=0$. What do you think?

You can factor out the $\sigma$ from the derivatives. Regarding your question, the spatial derivatives of the coordinate basis vectors $\hat{a}_{\theta}$ and $\hat{a}_z$ along the surface have components normal to the surface. This is what the pressure has to balance.

In anticipation of what we do next, what is the magnitude of the vector $\hat{a}_{\theta}\times \hat{a}_z$?

 

[member 428835]

From your previous post we have $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}\implies \\
\hat{a}_\theta \times \hat{a}_z = \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right) \times \left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
=-R \frac{\partial R}{\partial z}\hat z-\frac{\partial R}{\partial \theta}\hat \theta+R\hat r \implies\\
||\hat{a}_\theta \times \hat{a}_z|| = \sqrt{\left( R \frac{\partial R}{\partial z} \right)^2+\left( \frac{\partial R}{\partial \theta} \right)^2+R^2}$$
Geometrically this would be the area of a window along the surface. Is this what you're looking for?

 

[Chestermiller]

From your previous post we have $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}\implies \\
\hat{a}_\theta \times \hat{a}_z = \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right) \times \left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
=-R \frac{\partial R}{\partial z}\hat z-\frac{\partial R}{\partial \theta}\hat \theta+R\hat r \implies\\
||\hat{a}_\theta \times \hat{a}_z|| = \sqrt{\left( R \frac{\partial R}{\partial z} \right)^2+\left( \frac{\partial R}{\partial \theta} \right)^2+R^2}$$
Geometrically this would be the area of a window along the surface. Is this what you're looking for?

It's also equal to $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$. Can you show this?

 

[member 428835]

It's also equal to $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$. Can you show this?

I don't think so. I know we've seen that show up a bunch in the denominators. It also looks like a Jacobian, which relates transformed area. How would you show it, or do you have any suggestions on how to start me out?

 

[Chestermiller]

I don't think so. I know we've seen that show up a bunch in the denominators. It also looks like a Jacobian, which relates transformed area. How would you show it, or do you have any suggestions on how to start me out?

Here's a hint. The cross product of two vectors is equal to the magnitudes of the two vectors times the sine of the angle between them. The cosine of the angle between two vectors is equal to their dot product divided by the magnitudes of the two vectors. sin^2=1-cos^2

 

[member 428835]

Here's a hint. The cross product of two vectors is equal to the magnitudes of the two vectors times the sine of the angle between them. The cosine of the angle between two vectors is equal to their dot product divided by the magnitudes of the two vectors. sin^2=1-cos^2

Beautiful. We have $$|\hat a_\theta \times \hat a_z| = |\hat a_\theta|| \hat a_z|\sin\theta\\
= |\hat a_\theta|| \hat a_z|\sqrt{1-\cos^2\theta}\\
=|\hat a_\theta|| \hat a_z|\sqrt{1-\left(\frac{\hat a_\theta\cdot\hat a_z}{|\hat a_\theta|| \hat a_z|}\right)^2}\\
=\sqrt{(|\hat a_\theta|| \hat a_z|)^2-(\hat a_\theta\cdot \hat a_z)^2}\\
=\sqrt{\sqrt{\hat a_\theta\cdot\hat a_\theta}^2\sqrt{\hat a_z\cdot\hat a_z}^2-(\hat a_\theta\cdot \hat a_z)^2}\\
=\sqrt{g_{\theta\theta}g_{zz}-g_{\theta z}^2}$$

 

[Chestermiller]

OK. So, now going back to post #72, we have:
$$\sigma \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\sigma \frac{\partial}{\partial \theta}\left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p\hat{N}= \hat{0}$$where $\hat{N}=\hat{a}_{\theta}\times \hat{a}_z$. The next step is to determine the pressure p by dotting the equation with $\hat{N}$ to obtain:
$$\sigma \hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)+
\sigma \hat{N}\centerdot \frac{\partial}{\partial \theta}\left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$
See if you can show that this is the same as:
$$\sigma \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\sigma \left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$

 

[member 428835]

See if you can show that this is the same as:
$$\sigma \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\sigma \left( \frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}} \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}+
p(g_{zz}g_{\theta\theta}-g_{\theta z}^2)= 0$$

I have no clue how you did this. What I'm thinking, though it doesn't seem applicable here, is if you rotate a vector $\vec a$ by a given amount and dot it with $\vec b$ it will be the same as if you rotate $\vec b$ by the given amount and dot it with $\vec a$. But the derivatives aren't rotations. Honestly, to me this looks like dark magic.

 

[Chestermiller]

What is the dot product of N with the expressions in parenthesis?

 

[member 428835]

From the second to last equation in post 79, I'll consider the far left term (dropping the constant $\sigma$). The product rule implies
$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)=\frac{\partial}{\partial z}\left[\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\right] -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)$$
By the definition of $\hat N$ we have
$$\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right) = (\hat a_\theta \times \hat a_z)\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\\
=\left(\frac{g_{\theta\theta} (\hat a_\theta \times \hat a_z)\centerdot \hat a_z-g_{\theta z}(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right).
$$
Notice $(\hat a_\theta \times \hat a_z)\centerdot \hat a_z = 0$ and $(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta = 0$ since the dot product of two orthogonal vectors is zero. Thus the product inside the $z$ derivative from the first equation is zero, and we are left with

$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)= -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)
$$
Then I agree with your final equation in post 79 except that I think the two $\sigma$ terms are off by a sign. What do you think?

 

[Chestermiller]

From the second to last equation in post 79, I'll consider the far left term (dropping the constant $\sigma$). The product rule implies
$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)=\frac{\partial}{\partial z}\left[\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\right] -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)$$
By the definition of $\hat N$ we have
$$\hat{N}\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right) = (\hat a_\theta \times \hat a_z)\centerdot\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)\\
=\left(\frac{g_{\theta\theta} (\hat a_\theta \times \hat a_z)\centerdot \hat a_z-g_{\theta z}(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right).
$$
Notice $(\hat a_\theta \times \hat a_z)\centerdot \hat a_z = 0$ and $(\hat a_\theta \times \hat a_z)\centerdot\hat a_\theta = 0$ since the dot product of two orthogonal vectors is zero. Thus the product inside the $z$ derivative from the first equation is zero, and we are left with

$$\frac{\partial \hat{N}}{\partial z} \centerdot \left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)= -\hat{N}\centerdot \frac{\partial}{\partial z}\left(\frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}\right)
$$
Then I agree with your final equation in post 79 except that I think the two $\sigma$ terms are off by a sign. What do you think?

I don't think so, but we can check that later. So, from all this, if you do all the arithmetic, what equation do you get for the pressure p? We can compare notes.

 

[member 428835]

I get (using the negative result I got) $$p = \frac{\sigma}{\left(g_{zz}g_{\theta\theta}-g_{\theta z}^2\right)^{3/2}} \left[ \left(g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\left( g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}\right]
$$
Nothing special here except simple algebra.

 

[Chestermiller]

I get (using the negative result I got) $$p = \frac{\sigma}{\left(g_{zz}g_{\theta\theta}-g_{\theta z}^2\right)^{3/2}} \left[ \left(g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta\right)\centerdot \frac{\partial \hat{N}}{\partial z}+
\left( g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z \right)\centerdot \frac{\partial \hat{N}}{\partial \theta}\right]
$$
Nothing special here except simple algebra.

I meant to flesh out N, and do the dot products.

 

[member 428835]

I meant to flesh out N, and do the dot products.

hahahahhaha I thought so! To be clear, we're talking about taking the coordinates back into $z$ and $r$, right? I'll wait to crunch the numbers until you confirm. I know sometimes you have tricks up your sleeve.

 

[Chestermiller]

hahahahhaha I thought so! To be clear, we're talking about taking the coordinates back into $z$ and $r$, right? I'll wait to crunch the numbers until you confirm. I know sometimes you have tricks up your sleeve.

I'll try to write out my result so we can compare.

 

[Chestermiller]

OK, here's what I got for the pressure (if I did the arithmetic correctly). I corrected the sign error that you pointed out. Please check my results:

$$p=\frac{1}{R\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{1/2}}-\frac{\left[1+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]\frac{\partial^2 R}{\partial z^2}-2\frac{\partial R}{\partial z}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)\frac{1}{R}\frac{\partial ^2R}{\partial \theta \partial z}+\left[1+\left(\frac{\partial R}{\partial z}\right)\right]^2\frac{1}{R^2}\frac{\partial^2R}{\partial \theta^2}-\frac{1}{R}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2}{\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{3/2}}$$

 

[member 428835]

OK, here's what I got for the pressure (if I did the arithmetic correctly). I corrected the sign error that you pointed out. Please check my results:

$$p=\frac{1}{R\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{1/2}}-\frac{\left[1+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]\frac{\partial^2 R}{\partial z^2}-2\frac{\partial R}{\partial z}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)\frac{1}{R}\frac{\partial ^2R}{\partial \theta \partial z}+\left[1+\left(\frac{\partial R}{\partial z}\right)\right]^2\frac{1}{R^2}\frac{\partial^2R}{\partial \theta^2}-\frac{1}{R}\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2}{\left[1+\left[\frac{\partial R}{\partial z}\right)^2+\left(\frac{1}{R}\frac{\partial R}{\partial \theta}\right)^2\right]^{3/2}}$$

Hmmmm I'm not getting exactly this. Did you compute this result in Mathematica? If so I can send you my notebook. I plugged your result and mine in Mathematica and I am not getting zero when fully simplifying the difference of them both.

If you have Mathematica, let me know and I'll send you my notebook. If you don't have Mathematica, can we agree on the following components:

where $N_z = \partial_z\hat N$ and $N_\theta = \partial_\theta\hat N$. Note vectors are denoted as $\{a,b,c\} \equiv a\hat r + b\hat \theta + c\hat z$. Everything else is self explanatory I think.

 

[Chestermiller]

Hmmmm I'm not getting exactly this. Did you compute this result in Mathematica? If so I can send you my notebook. I plugged your result and mine in Mathematica and I am not getting zero when fully simplifying the difference of them both.

If you have Mathematica, let me know and I'll send you my notebook. If you don't have Mathematica, can we agree on the following components:View attachment 220353
where $N_z = \partial_z\hat N$ and $N_\theta = \partial_\theta\hat N$. Note vectors are denoted as $\{a,b,c\} \equiv a\hat r + b\hat \theta + c\hat z$. Everything else is self explanatory I think.

I confirm all those components.

 

[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?