Partial differentiation for dimensionless continuity equation

[psychlone]

Hi,

 

I start with the continuity equation, for a CFD application.

 

∂u/∂x+∂v/∂y=0 (1)

 

I’m interested in substituting the following (2)(3)(4)(5) into the above equation (1) to produce a non-dimensional continuity equation.

 

u = (UKRe/ρR) (2)

v = ((UK〖Re〗^(1⁄2))/ρR) (3)

 

x=ϕR (4)

y=ηδ (5)

 

which produces;

 

∂/∂(ϕR) (UKRe/ρR)+∂/∂(ηδ) ((UK〖Re〗^(1⁄2))/ρR)=0 (6)

 

∂/∂ϕ (UKRe/ρR).∂/∂R (UKRe/ρR)+∂/∂η ((UK〖Re〗^(1⁄2))/ρR).∂/∂δ ((UK〖Re〗^(1⁄2))/ρR)=0 (7)

 

My question is; when I do the substitution for x & y, and manipulating equation (6) would I yield equation (7)?

 

For completeness I’ve listed all the variables and constants used.

 

Where:

u = velocity (x-direction, vector)

v = velocity (y-direction, vector)

U = Velocity (x direction Dimensionless)

R = Radius

K = Constant

Re = Reynolds Number

Φ= Angle (Radians/ non-dimensional)

δ = film thickness (vector)

η = Dimensionless coordinate normal to the surface

ρ=density

 

Thanks.

Edited December 19, 2008 by psychlone
grammer

[Bignose]

1) This forum used LaTex, which will make it a little easier to put in your equations. Just use the [_math_] and the [_/math_] tags (remove the underscores _'s). If you quote my reply here, you'll see the math tags below.

 

2) When you substitute in dimensionless variables, you just use the normal rules of differentiation of constants.

 

I.e. let [math]x'[/math] be the dimensionless x variable. Something like:

 

[math]x' = \frac{x}{L}[/math] where L is some characteristic length.

 

Rearrange that and you get:

 

[math]x = Lx'[/math]

 

Now, when you take a differential of x, dx, you put in the substitution

 

[math] dx = d(Lx') = Ldx' [/math]. The L comes out because it is a constant. And so on. You can do this for any variable you made dimensionless.

[psychlone]

Thanks for the tip on entering equations and for your reply, it is much appreciated.

 

To the dimensionless variables;

 

Then; x=[math]\phi[/math]R

[math]\phi=const=\frac{dx}{dR}[/math]

 

That makes sense.

 

the solution I have to the continuity uses a partial differentiation with respect to dimensionless variables.

 

Could I do this;

 

[math]\frac{\partial u}{\partial x} = \frac{\partial}{\partial\left(\frac{x}{R}\right)}\left(\frac{UKRe}{\rho R}\right) = \frac{\partial}{\partial\phi}\left(\frac{UKRe}{\rho R}\right)[/math].

 

Where I take the derivative with respect to [math]x[/math]. and divide it through by [math]R[/math].

 

Thanks.

Edited December 19, 2008 by psychlone
multiple post merged

[Bignose]

Well,

 

1) Firstly, there is a LaTeX error in your post so I am not sure of what you are trying to do

 

but,

 

2) Secondly, I guess I am also unsure of what you are trying to do in the first post. Usually, the velocity is nondimensionalized by using a characteristic velocity. Such as the average entrance velocity, or the velocity of a moving boundary. Usually the Reynolds number isn't part of the nondimensionalizing of the velocity, because the Reynolds number contains a velocity itself. Sometimes a characteristic velocity is found by taking a characteristic length and dividing it by a characteristic time, like the stroke of a piston divided by the cycle time of the piston.

 

For an incompressible flow, the dimensionless continuity equation should look like dimensional continuity equation.

 

The Navier-Stokes equation will have dimensionless numbers in them that are the result of the nondimensionalization. The Reynolds number will be in front of the viscous diffusion term, the Euler number will be in front of the pressure term, the Froude number will be in front of the gravity terms, etc.

 

I am also confused by your use of combining x and R and calling R a radius. If you are using x & y, and the form of the continuity equation in your first equation, then you are using Cartesian coordinates -- there shouldn't be a "radius" that is proportional to a coordinate direction in this case. If there is a radius proportional to a coordinate direction, then you need to use something like cylindrical or spherical or maybe even generalized coordinates with an axisymmetry -- it depends on the application. Perhaps if you start from the top, I can try to point you in a better direction. But at the moment, I am pretty confused about what you are trying to do.

[psychlone]

Reply to 1). my apologies I had to edit my text to correct the LaTex errors, But my post should be reading math symbols now.

 

Reply to 2). I didn't include the characteristic velocity variable when multiplied by dimensional velocity yields a non-dimensional velocity.

 

where:

 

[math] u = \frac{U}{u_{r}}[/math]

 

[math]u_{r}=\frac{K Re}{\rho R}[/math]

 

 

And I agree, based on the Cartesian coordinate system the dimensional and non dimensional continuity should be the same.

 

And Your right, I've over looked building the equation with respect to the correct the coordinate system, which should be cylindrical or polar. This something I need to focus. This has probably the stumbling block that has prevented me from going any further.

 

Thanks very much for your help it's greatly appreciate.

 

 

And where did you study Fluid Mechanics I'm impressed with your insight. You can ignore this question if you like.

Edited December 19, 2008 by psychlone

[Bignose]

where:

 

[math] u = \frac{U}{u_{r}}[/math]

 

[math]u_{r}=\frac{K Re}{\rho R}[/math]

 

 

OK, but here's the question I have. What velocity are you using in your Reynolds number here? And why wouldn't you use that same velocity that you are using in the Reynolds number to nondimensionalize the velocity?

 

And furthermore, the Reynolds number comes out of the momentum equation -- it isn't the result of using it in the definition of the characteristic velocity -- it arises naturally from the process of nondimensionalization of the Navier Stokes equation.

 

I don't like giving out too much personal information on the Internet, but I will say that I attended two public schools for engineering.

 

Thanks for the tip on entering equations and for your reply, it is much appreciated.

 

To the dimensionless variables;

 

Then; x=[math]\phi[/math]R

[math]\phi=const=\frac{dx}{dR}[/math]

 

That makes sense.

 

the solution I have to the continuity uses a partial differentiation with respect to dimensionless variables.

 

Could I do this;

 

[math]\frac{\partial u}{\partial x} = \frac{\partial}{\partial\left(\frac{x}{R}\right)}\left(\frac{UKRe}{\rho R}\right) = \frac{\partial}{\partial\phi}\left(\frac{UKRe}{\rho R}\right)[/math].

 

Where I take the derivative with respect to [math]x[/math]. and divide it through by [math]R[/math].

 

Thanks.

 

This isn't right:

[math]\frac{\partial u}{\partial x} = \frac{\partial}{\partial\left(\frac{x}{R}\right)}\left(\frac{UKRe}{\rho R}\right) = \frac{\partial}{\partial\phi}\left(\frac{UKRe}{\rho R}\right)[/math].

 

Because you are taking a derivative with respect to a constant -- that will always be equal to zero.

 

----

 

If you want some help with this, I can show you the steps to generate the Reynolds number. There are actually 2 or 3 different ways to generate it, and it is important to know them all, in my opinion.

 

An excellent resource on nondimensionalizing is in the book Fundamentals of Fluid Mechanics by Munson, Young, and Okiishi.

[psychlone]

I'm using the velocity in the x - direction [math]\left(u\right)[/math].

As I’m solving for the first term in the continuity equation [math]\frac{\partial u}{\partial x}[/math].

Thus, your right, the velocity in Reynolds number with cancels with the [math] u [/math] variable term.

I'm going about the problem the wrong way, doing strictly substitution across different coordinate systems, which can be done by focusing on first is using the Polar form of the continuity equation and then the second step is to the substitute the vectors with scalars.

Thanks for your time, it is very much appreciated.

[Bignose]

I'm going about the problem the wrong way, doing strictly substitution across different coordinate systems, which can be done by focusing on first is using the Polar form of the continuity equation and then the second step is to the substitute the vectors with scalars.

 

OK, I'm just going to start clean here and show you how I learn how to do this, because I still don't think that you have a good idea. Phrases like "substitute the vectors for scalars" seems quite wrong. I think I know what you mean, but the phrasing of those words is very inexact, because you can't just replace a scalar for a vector -- a quantity is a vector, has to remain a vector, otherwise it isn't the same property. For example, if I say 10 mph due North West, that is a vector because it has a magnitude and a direction. I can't just "substitute" in 10 mph. Because that isn't a vector; it doesn't have a direction. In this case, it is just a speed, a different -- though related -- concept from velocity.

 

I suspect that what you meant was to "solve for the components of the velocity" in that you write out the 2 or 3 equations for the components of the velocity vector. That is, the equation for [math]v_x, v_y, v_z[/math] in Cartesian coordinates, or [math]v_r, v_{\theta}, v_z[/math] in cylindrical coordinates and so on for the other coordinates. It is a small-thing, but the words you use have exact meanings especially in math and physics, so you have to chose them carefully

 

Now, on to nondimensionalizing the Navier-Stokes Equations. Now, this can either be done in their vector form, or in their individual component form -- the results are the same. I'm going to do it in the vector form.

 

As a nomenclature note, unprimed variables have dimensions, primed variables are dimensionless, and I use the symbol [=] to mean "has dimensions of" For example velocity, v [=] L/T which means length per time.

 

The N-S eqns (I'm going to drop the pressure and gravity/body force terms, as they make more dimensionless groups -- ideally, for practice, you should probably put them back in and discover them in the same way I will post below).

 

[math]\rho \frac{\partial{\mathbf{v}}}{\partial{t}} + \rho \mathbf{v}\cdot \nabla \mathbf{v} = \mu \nabla^2 \mathbf{v} [/math]

 

Now, the variables are:

 

density = [math]\rho[/math] [=] [math] M/L^3[/math]

velocity = [math]\mathbf{v}[/math] [=] [math]L/T[/math]

time = [math]t[/math] [=] [math]T[/math]

viscosity = [math]\mu[/math] [=] [math]M/(LT)[/math]

 

on the right side of the [=]'s, the M stands for mass, L for length, T for time -- viscosity has units of mass per length per time.

 

To nondimensionalize this, we need some characteristic values. In most cases, there are going to be some pretty obvious ones. The radius or diameter of the pipe is an obvious characteristic length. The average inlet velocity is a an obvious characteristic velocity. Or, the velocity of a moving boundary is an obvious characteristic velocity.

 

Whatever the are, let's just assume that they exist. So, let [math]l_c[/math] be a characteristic length and let [math]v_c[/math] be a characteristic velocity.

 

Note that with a characteristic velocity, we can make a characteristic time, [math]t_c = \frac{l_c}{v_c}[/math]

 

Furthermore let the dimensionless variables be defined as:

[math]\mathbf{v}' = \frac{\mathbf{v}}{v_c}[/math]

 

[math]t' = \frac{t v_c}{l_c}[/math]

 

[math]\nabla' = l_c \nabla[/math]

 

We're going to leave the density and viscosity alone. In all honesty, this is because I know the result we are looking for. But, you could actually nondimensionalize them too, if you found a characteristic mass. This would be important in a compressible flow, a characteristic mass could come from a density as a reference temperature and pressure, for example.

 

It is important to note that there is a nondimensional gradient operator, because the dimensional gradient operator has units of inverse length.

 

I like to re-write those equations above so that they are in a form easy to insert into the equaiton:

 

[math] v_c \mathbf{v}' = \mathbf{v}[/math]

 

[math]\frac{t' l_c}{v_c} = t[/math]

 

[math]\frac{1}{l_c} \nabla' = \nabla[/math]

 

Now, just plug these into the N-S equation above:

 

[math]\rho \frac{\partial{v_c \mathbf{v}'}}{\partial{\frac{t' l_c}{v_c}}} + \rho v_c \mathbf{v}'\cdot \frac{1}{l_c} \nabla' v_c \mathbf{v}' = \mu (\frac{1}{l_c} \nabla')^2 v_c \mathbf{v}' [/math]

 

now, let's collect all the constants together in front of each term:

 

[math]\rho \frac{v_c^2}{l_c} \frac{\partial{\mathbf{v}'}}{\partial{t'}} + \rho \frac{v_c^2}{l_c} \mathbf{v}' \cdot \nabla' \mathbf{v}' = \frac{\mu v_c}{l_c^2} \nabla'^2 \mathbf{v}' [/math]

 

Divide through by [math]\rho \frac{v_c^2}{l_c}[/math] and you get:

 

[math]\frac{\partial{\mathbf{v}'}}{\partial{t'}} + \mathbf{v}' \cdot \nabla' \mathbf{v}' = \frac{\mu}{l_c v_c \rho} \nabla'^2 \mathbf{v}' [/math]

 

And finally, note that term in front of [math]\nabla'^2 \mathbf{v}' [/math]..

 

It is a Reynolds number. A velocity times a length times a density divided by viscosity is a Reynolds number.

 

[math]Re=\frac{l_c v_c \rho}{\mu}[/math] so

 

[math]\frac{\partial{\mathbf{v}'}}{\partial{t'}} + \mathbf{v}' \cdot \nabla' \mathbf{v}' = \frac{1}{Re} \nabla'^2 \mathbf{v}' [/math] <--- the dimensionless Navier-Stokes equation.

 

The Reynolds number comes out completely naturally as a result of the nondimensionalization process, and since I did it with the vector equation, it is completely coordinate system independent.

 

Like I said, you should put the pressure and the gravity terms back into the N-S equations and repeat the above to discover that the Froude number

 

[math]Fr = \frac{v_c^2}{g l_c}[/math]

 

and the Euler number

 

[math]Eu = \frac{\Delta p_c}{\rho v_c^2}[/math] (the [math]\Delta p_c[/math] would be a characteristic pressure drop)

 

comes out just as naturally as the Reynolds number comes out. In the same way, the Nusselt number comes out of the equations of heat transfer, and many of the other famous dimensionless numbers.

 

-----

 

Now, I mentioned that there is another way to come up with these numbers, and that is the very powerful Buckingham Pi Theorem. http://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem With it, you take all the variables in the equation, and you predict all the possible dimensionless numbers that can come out of the problem at hand. It won't show you where the dimensionless numbers arise in the equations, but it will alert you to which ones are relevant to the problem at hand.

 

I hope that this makes it all clearer. I do also hope that you will put in the gravity term and find the Froude number. I also hope that you will follow the same procedure on the continuity equation to find the dimensionless continuity equation. This can also be done with the energy equation -- really any of the equations of fluid mechanics.

[psychlone]

I've looked at my problem again, and the coordinates are in Cartesian form. The reason for this is the problem is;

 

The derivation is uses a Cartesian coordinate system for fluid flowing around the circumference of a cylinder. Where x, is not flow though a pipe but x is flow of flow of fluid parallel to the surface of the cylinder and y is always perpendicular to the surface of the cylinder, which is the thickness of the film.

 

2 dimensional only, looking only at a cross section of the cylinder.

My apologies once again I should have mentioned all this earlier.

 

The answer to the derivation;

 

[math]\frac{\partial U}{\partial \phi} - \frac{\eta}{\delta_{a}} \frac{d \delta_{a}} {d \phi} U \frac{\partial U}{\partial \eta} + \frac{1}{\delta_{a}} \frac{\partial V}{\partial \eta} [/math]

 

Where:

 

[math] u_{r} = \frac{K Re}{\rho R} [/math]

 

Where: [math] K = [/math] Fluid consistency index (Pascals sec) (SI uits)

 

[math] V = \frac{v Re}{u_{r}}[/math]

 

[math] \eta = \frac{y}{\delta (x)} = \frac{y Re}{R \delta_{a}}[/math]

 

*But if [math]\phi[/math] & [math]\eta[/math] are variables but are still dimensionless, taking a partial differential with respect to those variables has to be valid.

 

This derivation is from;

Laminar flow and heat transfer or non-Newtonian falling liquid on a horizontal tube with variable surface heat flux (D. Oudhadda & A. II Idrissi)

Int. Comm. Heat Transfer. Vol. 28. No 8, pp 1125- 1135, 2001.

 

What I meant with the comment "substituting scalars with vectors" was scalars are dimensionless but are taken as individual components in the direction of the vector.

Can a single characteristic velocity [math] u_{r} [/math] be used for both [math u [/math] & and [math] v [/math] velocity to convert them.

 

**Bearing in mind in the above Reynolds number term uses the [math] u [/math] velocity, which is the velocity in the direction parallel to the surface of the cylinder.

Edited December 20, 2008 by psychlone
multiple post merged

[Bignose]

OK,

 

I looked at your paper there.

 

1) K is a viscosity-like parameter. It is the factor in front of the power-law model for the shear stress in the non-Newtonian fluid they used. K has units so that when it is multiplied by the shear rate to a power, the correct units of shear stress come out

 

2) [math]Re_m[/math] is a modified Reynolds number. Modified in several ways. Firstly, they use the mass flux in place of a characteristic velocity. That's what the [math]\Gamma[/math] stands for. Secondly, because of the power law model for the shear stress, the K part has to be raised to a the correct power so that the Reynolds number becomes dimensionless.

 

Then, using this mass flux, they define a characteristic velocity, [math]u_r[/math]

 

All this is fine with what I said above -- because like I said sometimes it takes some manipulation to get the characteristic variables you want. This one looks a lot different because of the different model for the shear stress.

 

The Navier-Stokes equations are only for Newtonian fluids, and Newtonian fluids obey the shear stress rule:

 

[math]\mathbf{\tau} = -\mu\mathbf{\dot{\gamma}}[/math]

 

where

 

[math]\mathbf{\dot{\gamma}} = \nabla\mathbf{v} + (\nabla\mathbf{v})^T[/math] is the rate of strain tensor.

 

For the power law fluid in your paper,

 

[math]\mathbf{\tau} = -K(\mathbf{\dot{\gamma}})^n[/math] where n is some constant.

 

Using this as the shear-stress in the momentum conservation equations yields equations that are non longer Navier Stokes equations. N-S assumes Newtonian flow.

 

3) That paper isn't too terribly clear about what coordinates it is using. It is a film running down the inside of a spherical shell, but it looks like the authors are trying to use Cartesian coordinates to describe the flow. This is where the extra terms in the continuity equation come in. Because the film thickness, [math]\delta[/math] changes, there are extra terms that must account for the expansion and contraction of the film. In particular, note how the [math]\eta[/math] is a function of both x & y because of the coordinate system chosen. When you use the chain rule, that's why you get derivatives of U with respect to both the nondimnsionalized x & y -- because the nondimensionalized y, the [math]\eta[/math] is a function of both x & y. It really isn't very clear.

 

Then, on top of it, I am really confused because you have some evaporation but I don't seem a mass-loss term in the boundary conditions.

 

It might be worthwhile looking for one of those author's thesis document or a equivalent where there might be more detail listed. I think that I personally would have tried to use spherical coordinates, but that may have had its own set of issues that complicate things. I do know that if I were refereeing that paper, I would have asked the authors for more detail.

 

4) Here's my final thoughts. Have you done the same problem, just with a Newtonian fluid? Because, in things like this, you really want to learn to walk before you run. Especially if you aren't experienced with non-Newtonian fluids, is you aren't experienced with CFD, etc. It may even be worthwhile to go all the way back and re-solve a falling film problem of just a Newtonian fluid flowing down an inclined plane. Then do a power-law fluid flowing down an inclined plane. Then do a Newtonian fluid film flowing around a sphere like in the problem, then the power low fluid, etc.

 

Introduce one complexity at a time until you build up to the problem at hand. If you try to put too much into it all at once, you'll never know where your mistake is. Or even worse, you'll make two mistakes, and they'll almost cancel each other out until you find and fix one of the them and then things will get a lot worse and you won't understand because you thought you fixed something. Believe me, I've been there with fluid dynamics and CFD problems before. It really is best to start as simple as possible and slowly add complexity to the problem. Even if you think you have a good handle on what to do with the complex phenomena. This suggestion even includes CFD solution methods. Don't jump into a complex solution method -- start with the really old basics and slowly put in the more complex and accurate ones. Again, if you put in too many changes at once you won't know what change or changes caused the error.

[psychlone]

you haven't answered my question

[Bignose]

you haven't answered my question

 

Dude, I downloaded and looked at the paper, I took the time to explain what I saw. And I don't see any question in the last several posts!

 

I think (and I hope) you didn't meant it to come of that way, but your post sure reads awfully rude to someone who has volunteered a lot of time typing up equations trying to help you as best I can.

 

Now, I'm going to give you a second chance here and ask what question I didn't answer, and I'll do my best to answer it for you. But there aren't any guarantees. I'm not the author of the paper or anything, so I don't know what all the authors did in their paper.

[psychlone]

I don’t mean to be rude, and I’m sorry it come across like that & I greatly appreciate your time and effort and knowledge, but as much of a help you are your beginning to frustrating me.

My original question is;

*But if [math] \phi [/math] & [math] \eta [/math] are variables but are still dimensionless, taking a partial differential with respect to those variables has to be valid.

Because you said that taking partial derivative to these variable are similar to taking the derivative of a constant.

 

I meant to back to you sooner with my qustion but I had to go do something.

 

Don't take this the wrong way but let's just forget about it.

[Bignose]

 

Then; x=[math]\phi[/math]R

[math]\phi=const=\frac{dx}{dR}[/math]

 

 

My statement about taking the derivative of a constant stems from this statement right here. If [math]\phi[/math] is a constant -- as you wrote in your equation -- then taking a derivative with respect to it really doesn't have any meaning, right? It's be like asking what is [math]\frac{df}{d2}[/math] you see?

 

Now, I think what happened here is that R is the constant, not [math]\phi[/math]. So, this statement above isn't right. But, this is what I was responding to. Taking the derivative with respect to a constant doesn't really have any meaning.

 

There is no reason to be frustrated, I'm just trying to help answer the questions that I see, that I know about. You shouldn't feel frustrated, because you could have gotten no help at all. Your feeling frustrated conveys some sense of entitlement -- like you are frustrated that you aren't getting the help you should be -- when instead on a free forum where people are volunteering their time you should probably be grateful that anyone is trying to help you at all. This is 100% a community of volunteers who spend their time helping others. We are not paid. Part of the frustration is that I don't think that you are being 100% clear in what you are asking. I am trying my best, but if you need different answers, you need to be clearer about exactly what you want. If I can help, I will try to help, but this isn't a guarantee. I may not be able to give you the answers you want, and I may give you wrong answers, because we all make mistakes. I may not be able to give you any answer at all. But to feel frustrated I think it really quite unfair to me and to the entire forum.

Edited December 21, 2008 by Bignose

[psychlone]

my apologies.