When to use which dimensionless number

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I'm sorry, I am not able to provide a summary as the conversation is too technical and involves equations.
  • #1
member 428835
Hi PF!

I've been reading about low gravity capillary driven flows, and no authors use Reynolds number when measuring importance of inertia in capillary driven flows. Instead most use the Ohnesorge number. Can someone explain why this is?

Thanks!
 
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  • #2
Did you google ?
 
  • #3
BvU said:
Did you google ?
I did. The Ohnesorge number is evidently like a Reynolds number but with a capillary length scale. What would cause someone to think of this?
 
  • #4
Significance of surface tension in capillary flow
 
  • #5
BvU said:
Significance of surface tension in capillary flow
But there doesn't seem to be anything intrinsically wrong with Reynolds number. Sure, it doesn't measure surface tension, but why isn't it really ever used for low Bond number capillary flows?
 
  • #6
@Chestermiller : you know of someone in PF with specific expertise here ? Unfortunately for Josh, mine is just hearsay :sorry: .
 
  • #7
BvU said:
@Chestermiller : you know of someone in PF with specific expertise here ? Unfortunately for Josh, mine is just hearsay :sorry: .
Very often in systems like this, we start out with the differential equations and use a systematic approach developed by Hellums and Churchill to reduce the equations to dimensionless form. The key dimensionless groups automatically emerge from this methodology (with some small leeway on selection of the groups). As first year graduate students at Michigan, we were taught this simple methodology by Stu Churchill back in 1963 (at that time, he was chairman of the ChE Dept.). It has served me well over the years.
 
  • #8
Thanks, Chet !

Also consulted with my roommate (from my point of view an expert...) who knows all these numbers, but he didn't come much further than: we use it in droplet formation to find correlations.

My own googling didn't get me far, but I did like the overview lemma.

As a physicist, when I 'm in a bad mood, I am inclined to think that these chemical engineers make up these dimensionless numbers as much as they can for two reasons: 1. as a claim to fame and immortality and 2. to make parity plots (preferably log-log) where at least three of these numbers appear with fractional exponents. :wink: But only when I'm in a bad mood, of course. And my roommate is a mathematician.
 
  • #9
hahahahaha this is a funny point! :oldlaugh: Ok then, maybe it's simply from scaling the equations! Thanks for your guys' help!
 
  • #10
Well, the Ohnesorge number is related to the Weber number and the Reynolds number, so under certain conditions (##We\approx 1##) it is effectively equivalent to Reynolds number.

The greater point here is that the choice of dimensionless group depends on the phenomena being studied at the time. The point made by @Chestermiller is important in that these dimensionless groups fall out of the governing equations (most importantly the Navier-Stokes equations) when you make them dimensionless based on scales that are important to the problem you are studying. However, what those scales are is up to you to define and will change from one class of problems to the next. Alternatively, one can carry out a dimensional analysis of the problem (e.g. by the Buckingham Pi Theorem) and those groups will fall out based on the various combinations of important parameters to the problem.

The bottom line is that different physics are dominant for different problems, and your choice of dimensionless groupings must reflect that.
 
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  • #11
Josh,

Why don't you provide the dimensional equations for your specific problem, and I'll illustrate the procedure.

Chet
 
  • #12
Chestermiller said:
Josh,

Why don't you provide the dimensional equations for your specific problem, and I'll illustrate the procedure.

Chet
The problem is fluid flowing down a wedge, though I think I understand the scaling, if that's what you're referring to? If ##z## is the direction of flow, ##y## is the coordinate going into the wedge, and ##x## goes from the corner upwards, ##\alpha## the half-corner angle of the wedge, and ##L## a characteristic length (in ##z## direction) and ##H## characteristic height (in ##x## direction) then scales look like
$$z \sim L\\
x \sim H\\
y \sim H\tan\alpha.$$
In this problem the Bond number is very small, and gravity can be neglected. Then Young-Laplace governs pressure (we assume negligible surface stress, since air is above the fluid) and the transverse curvature is assumed much greater than axial, reducing pressure equation to:
$$\Delta P = \sigma\frac{1}{R}\implies\\
P \sim \frac{\sigma}{fH}$$
where ##f = (\cos\theta/\sin\alpha-1)^{-1}## is a geometric function such that ##fH=R## where ##R## is the radius of curvature in the ##x-y## plane and ##\theta## is static contact angle of fluid. Then scaling the ##z## component of Navier Stokes yields to order ##\epsilon \equiv H/L##:
$$P_z = \sin^2\alpha\partial^2_xw+\cos^2\alpha\partial^2_yw.$$
It makes perfect sense to me that the Ohnesorge number emerges with the inertial terms (not included here for simplicity). However, to get a Reynolds number you'd have the Weber number like boneh3ad said. It's just interesting to me that this fluids problem doesn't really require a Reynolds number. What do you both think?
 
  • #13
I think I'd still like to see the formulation of the differential equations and boundary conditions.
 
  • #14
Chestermiller said:
I think I'd still like to see the formulation of the differential equations and boundary conditions.
Before I continue, I forgot a very important scale: ##w##. If we assume pressure balances viscosity, then dimensional constant density NS yields in ##z## component
$$\partial_zP = \mu \nabla^2 w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2}+\frac{1}{H^2\tan^2\alpha}+\frac{1}{L^2}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\tan^2\alpha + 1}{H^2\tan^2\alpha}\right)w\implies\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\sec^2\alpha}{H^2\tan^2\alpha}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2\sin^2\alpha}\right)w\implies\\
w \sim \frac{\sigma H\sin^2\alpha}{ f L \mu}\implies\\
w \sim \frac{\epsilon \sigma \sin^2\alpha}{ f \mu}.
$$
Notice I neglected ##1/L^2## term since ##\epsilon = H^2/L^2\ll 1## (slender column). Also, I would guess time scales as ##t \sim L/w##, though I'm not sure why I'm inclined to use ##w## rather than ##u,v##. Most of the flow is ##z## directed; is that good enough reason? Anyways, the formulation of the differential equations, assuming we start with dimensional NS and constant density, would be (for ##z## component, though I could do the other two if you'd like as well)

$$
\rho\frac{D w}{Dt} = \partial_zP+\mu \nabla^2w\implies\\
\frac{\rho W}{L/W}\frac{D w}{Dt} = \frac{\sigma}{HfL}\partial_zP+\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w\implies\\
\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\frac{HfL}{\sigma}\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w
$$

but we already balanced pressure and viscosity, so we know both terms on the RHS are now ##O(1)##, thus we have

$$\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon \sigma \sin^2\alpha}{ f \mu}\right)^2\frac{\rho }{1}\frac{Hf}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{\sigma \rho H}{\mu^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{1}{Oh^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\$$
and ##Oh \equiv \mu/\sqrt{\rho \sigma H}##. How does this look?
 
  • #15
And boundary conditions...oh gosh. So zero slip on the wedge, zero stress along the free surface. Hmmmm physically aren't those it?
 
  • #16
joshmccraney said:
The problem is fluid flowing down a wedge, though I think I understand the scaling, if that's what you're referring to? If ##z## is the direction of flow, ##y## is the coordinate going into the wedge, and ##x## goes from the corner upwards, ##\alpha## the half-corner angle of the wedge, and ##L## a characteristic length (in ##z## direction) and ##H## characteristic height (in ##x## direction) then scales look like
$$z \sim L\\
x \sim H\\
y \sim H\tan\alpha.$$
In this problem the Bond number is very small, and gravity can be neglected. Then Young-Laplace governs pressure (we assume negligible surface stress, since air is above the fluid) and the transverse curvature is assumed much greater than axial, reducing pressure equation to:
$$\Delta P = \sigma\frac{1}{R}\implies\\
P \sim \frac{\sigma}{fH}$$
where ##f = (\cos\theta/\sin\alpha-1)^{-1}## is a geometric function such that ##fH=R## where ##R## is the radius of curvature in the ##x-y## plane and ##\theta## is static contact angle of fluid. Then scaling the ##z## component of Navier Stokes yields to order ##\epsilon \equiv H/L##:
$$P_z = \sin^2\alpha\partial^2_xw+\cos^2\alpha\partial^2_yw.$$
It makes perfect sense to me that the Ohnesorge number emerges with the inertial terms (not included here for simplicity). However, to get a Reynolds number you'd have the Weber number like boneh3ad said. It's just interesting to me that this fluids problem doesn't really require a Reynolds number. What do you both think?
I don't understand your description of the physical problem. Can you please provide a diagram?
 
  • #17
joshmccraney said:
The problem is fluid flowing down a wedge,
<snip>
In this problem the Bond number is very small, and gravity can be neglected. <snip>

But what is driving the flow? An (external) applied pressure gradient? Marangoni effect? Something else?

Edit: hang on, I re-read your OP. Capillary driven flow. Got it.

Edit #2: Here's a reference you may find useful, especially sections 2.2 and 2.3: https://ac.els-cdn.com/S03019322060...t=1512499461_7ce3a4db15d949752b591ed73f002381
 
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  • #18
Chestermiller said:
I don't understand your description of the physical problem. Can you please provide a diagram?
Here is a schematic of the flow.
 

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  • #19
Andy Resnick said:
But what is driving the flow? An (external) applied pressure gradient? Marangoni effect? Something else?

Edit: hang on, I re-read your OP. Capillary driven flow. Got it.
Hi Andy.

If you understand the geometry, maybe you can provide a diagram before Josh does.
 
  • #20
joshmccraney said:
Here is a schematic of the flow.

Chestermiller said:
Hi Andy.

If you understand the geometry, maybe you can provide a diagram before Josh does.

This is a 'standard' problem- but not a simple one. There's at least one book and 1 NASA flight experiment that is devoted to this specific problem. Can't immediately find a PDF breaking down the governing equations, but I'm sure the OP can find a suitable reference.
 
  • #21
Andy Resnick said:
This is a 'standard' problem- but not a simple one. There's at least one book and 1 NASA flight experiment that is devoted to this specific problem. Can't immediately find a PDF breaking down the governing equations, but I'm sure the OP can find a suitable reference.
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 :oldbiggrin:
 
  • #22
joshmccraney said:
Here is a schematic of the flow.
I'm sorry Josh. I can't make sense out of this figure. I'm more confused than ever. Can you please try again?
 
  • #23
Chestermiller said:
I'm sorry Josh. I can't make sense out of this figure. I'm more confused than ever. Can you please try again?
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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  • #24
joshmccraney said:
Here's a sketch I drew a while ago. I probably should've used this earlier. The corner angle measures ##2\alpha## (not shown).
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?
 
  • #25
Chestermiller said:
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?
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?
 
  • #26
joshmccraney said:
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?
I don't know yet. I'm just starting to think about this problem.
 
  • #27
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.
 
  • #28
Chestermiller said:
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.
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.
 
  • #29
joshmccraney said:
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 :oldbiggrin:

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).

Maybe you've already seen this:
https://www.cambridge.org/core/jour...erior-corner/8C88D0772C50153457B155B8B1DDB029
 
  • #30
Andy Resnick said:
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
That's a good link for dimensionless numbers.

Andy Resnick said:
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).

Maybe you've already seen this:
https://www.cambridge.org/core/jour...erior-corner/8C88D0772C50153457B155B8B1DDB029
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 ;)
 
  • #31
joshmccraney said:
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.
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?
 
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  • #32
Chestermiller said:
OK Josh. In this spirit, this is how I would obtain the Young-Laplace equation for the free surface:
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?

Chestermiller said:
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.
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)?

Chestermiller said:
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?
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.
 
  • #33
joshmccraney said:
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?
Whenever the free surface has significant curvature, Young-Laplace needs to be used.
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)?
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?
 
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  • #34
Chestermiller said:
So, are you ready to move to the next step now?
Makes a lot of sense! Yes, I'm ready to see what's next!
 
  • #35
joshmccraney said:
Makes a lot of sense! Yes, I'm ready to see what's next!
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.
 
<H2>1. What is a dimensionless number?</H2><p>A dimensionless number is a mathematical quantity that expresses a relationship between physical variables without any units. It is typically used to compare different systems or processes and determine their similarities or differences.</p><H2>2. When should I use dimensionless numbers?</H2><p>Dimensionless numbers are useful when you want to compare different systems or processes that have similar physical properties. They can also be used to identify the dominant forces or parameters in a system and understand its behavior.</p><H2>3. What are some common dimensionless numbers used in science?</H2><p>Some common dimensionless numbers used in science include the Reynolds number, Froude number, Mach number, and Peclet number. These numbers are used in various fields such as fluid mechanics, thermodynamics, and aerodynamics.</p><H2>4. How do I know which dimensionless number to use?</H2><p>The choice of dimensionless number depends on the specific system or process you are studying. You should select a dimensionless number that represents the dominant forces or parameters in the system. It is also important to consider the physical properties and variables that are relevant to your analysis.</p><H2>5. Can dimensionless numbers be used in all scientific disciplines?</H2><p>Yes, dimensionless numbers can be used in various scientific disciplines such as physics, chemistry, engineering, and biology. They are a fundamental tool in understanding and analyzing complex systems and processes.</p>

1. What is a dimensionless number?

A dimensionless number is a mathematical quantity that expresses a relationship between physical variables without any units. It is typically used to compare different systems or processes and determine their similarities or differences.

2. When should I use dimensionless numbers?

Dimensionless numbers are useful when you want to compare different systems or processes that have similar physical properties. They can also be used to identify the dominant forces or parameters in a system and understand its behavior.

3. What are some common dimensionless numbers used in science?

Some common dimensionless numbers used in science include the Reynolds number, Froude number, Mach number, and Peclet number. These numbers are used in various fields such as fluid mechanics, thermodynamics, and aerodynamics.

4. How do I know which dimensionless number to use?

The choice of dimensionless number depends on the specific system or process you are studying. You should select a dimensionless number that represents the dominant forces or parameters in the system. It is also important to consider the physical properties and variables that are relevant to your analysis.

5. Can dimensionless numbers be used in all scientific disciplines?

Yes, dimensionless numbers can be used in various scientific disciplines such as physics, chemistry, engineering, and biology. They are a fundamental tool in understanding and analyzing complex systems and processes.

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