Navier-Stokes spherical discretization

lost.identity · April 14, 2010, 16:52

Hi,

I'm trying to discretize the Navier-Stokes equation written in the spherical form for a Finite Volume formulation. I'm having problems due to the 1/r^2 terms appearing in the equation.

I'm following the method of Patankar "Numerical heat transfer and fluid flow" where he integrates the equation for the control volume.

It is spherically symmetric so I'm only retaining the radial terms in my equations. Does anyone know how to do this?

Thanks.

ztdep · April 14, 2010, 22:56

use the control volume center coordiante to compute the 1/r^2.

lost.identity · April 15, 2010, 05:01

My problem is actually with integrating the equation.

I'm using Favre averaged version of the equation so I have the following convection term

(1/r^2)*d/dr(r^2*rho*u*u)

because u (the velocity) are functions of r I need to integrate this by parts. I tried multiplying the whole equation by r^2, but then this causes the same problem to other terms that doesn't have the (1/r^2) term.

I've tried searching through the literature to find any papers that does finite volume method on spherical Navier-Stokes equation and haven't been able to find any that have done the same thing as what I'm trying to do.

ztdep · April 15, 2010, 06:16

multiply r2 at both sides

lost.identity · April 15, 2010, 07:48

Thanks for the reply.

As I've already mentioned in my email, I've already tried that method. But the problem is when I multiply by r^2 on both sides I get terms like r^2*dp/dr, which again has to be integrated by parts.

But since p is generally a function of r this causes problems in the integration.

In addition I have more complex terms that arises due to Favre averaging such as r^2 d/dr(<rho*u''*u''>), where <> refers to Favre averages. I use the Boussinesq approximation to model this but still I get terms such as r*u, which again needs to be integrated by parts.

Surely, there's another way to do this?

ztdep · April 15, 2010, 18:59

nope do not integrate by part. take r^2 is a constant

lost.identity · April 16, 2010, 05:19

r^2 is not a constant, it arises by writing the Navier-Stokes equation in spherical form. r is one of the variable dimensions.

So when I'm integrating with respect to r, r^2 can't remain a constant.

agd · April 16, 2010, 08:57

I'm not sure I'm following what you are actually trying to do at this point. It sounds as if you are trying to integrate the full Navier-Stokes equations analytically. If you want to do a finite volume discretization, then the 1/r**2 term will be a part of what you discretize - in that case you will use the value at the cell center, as ztdep has suggested. Treat it just like any other term in the discretized equation.

lost.identity · April 16, 2010, 10:41

Sorry I should've explained it a bit more clearly

So I have the following 1-D Navier-Stokes equation written in spherical coordinates

$\frac{\partial\overline\rho{\tilde{u}}}{\partial{t}}+ \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\rho}{\tilde{u}\tilde{u}}\right)= -\frac{\partial\overline{p}}{\partial{r}} - \frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) + \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\tau_{rr}} \right)$

where

$\overline{\tau_{rr}} = {2\mu\frac{\partial{\tilde{u}}}{\partial{r}} -\frac{2}{3}\mu \frac{1}{r^2}\frac{\partial{r^2\tilde{u}}}{\partial{r}}}$

$\overline{\rho{u''}{u''}} = - {2\mu_t\frac{\partial{\tilde{u}}}{\partial{r}} +\frac{2}{3}\mu_t \frac{1}{r^2}\frac{\partial{r^2\tilde{u}}}{\partial{r}}} + \frac{2}{3}\overline{\rho}\tilde{k}$

Now what I want to do is integrate this momentum equation over the control volume and over time. Since it's 1-D I only consider a control volume around a single grid point P, the grid point East of point P is referred to as E and to the West is referred to as W.

Let's say I want to integrate the term

$\int_t^{t+\Delta{t}}\,\int_P^E\, \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\rho}{\tilde{u}\tilde{u}}\right) \, dr \, dt$

If the

term wasn't there this would have been straightforward, which is the case in the Cartesian coordinates.

So one thing I tried to do was to multiply the whole equation by

. But then I get terms like

$\int_t^{t+\Delta{t}}\,\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr\,dt$

which again gives me the same problem. This may be different to what's normally done in Control Volume analysis.

I'm trying to use the method of Patankar "Numerical Heat Transfer and Fluid Flow", 1980.

Thanks.

ztdep · April 16, 2010, 12:07

r^2 tem represent the element volume in the sperical coordinate system Dv. as in cartesian coordinate, it is 1 dx dy dz.

lost.identity · April 16, 2010, 13:41

Thanks, I think I sort of know what you're saying.

Are you saying that when I multiply by

, and integrate then terms such as

$\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr= \Delta{V}\left [ (\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P \right ]$

is this true for 1-D case such as mine? and would $\Delta{V}=1$ ?

or is $\Delta{V} = r_P^2\Delta{r}$

where

is the radial value at the mid-point of the control volume?

ztdep · April 16, 2010, 21:36

yes, that is what i mean/
by the way , how do you input the formular?

lost.identity · April 17, 2010, 17:06

THanks

Just want to confirm, you mean

$\Delta{V} = r_P^2\Delta{r}$

The site uses Latex formatting to type equations. If you click any of the equations you'd see the code.

lost.identity · April 18, 2010, 12:39

Actually shouldn't it be

$\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr = \Delta{V} \left [\frac {\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \right ]_e= \Delta{V} \left [ \frac{(\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P}{\Delta{r}} \right ]$

where $\Delta{V} = r_e^2\Delta{r}$ , P and E refers to the left and right hand side faces control volume respectively, and e is the midpoint of the control volume.

which would finally give

$\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr = r_e^2 \left[ (\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P \right ]$

April 14, 2010, 16:52	Navier-Stokes spherical discretization	#1
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Hi, I'm trying to discretize the Navier-Stokes equation written in the spherical form for a Finite Volume formulation. I'm having problems due to the 1/r^2 terms appearing in the equation. I'm following the method of Patankar "Numerical heat transfer and fluid flow" where he integrates the equation for the control volume. It is spherically symmetric so I'm only retaining the radial terms in my equations. Does anyone know how to do this? Thanks.

April 16, 2010, 10:41		#9
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Sorry I should've explained it a bit more clearly So I have the following 1-D Navier-Stokes equation written in spherical coordinates $\frac{\partial\overline\rho{\tilde{u}}}{\partial{t}}+ \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\rho}{\tilde{u}\tilde{u}}\right)= -\frac{\partial\overline{p}}{\partial{r}} - \frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) + \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\tau_{rr}} \right)$ where $\overline{\tau_{rr}} = {2\mu\frac{\partial{\tilde{u}}}{\partial{r}} -\frac{2}{3}\mu \frac{1}{r^2}\frac{\partial{r^2\tilde{u}}}{\partial{r}}}$ $\overline{\rho{u''}{u''}} = - {2\mu_t\frac{\partial{\tilde{u}}}{\partial{r}} +\frac{2}{3}\mu_t \frac{1}{r^2}\frac{\partial{r^2\tilde{u}}}{\partial{r}}} + \frac{2}{3}\overline{\rho}\tilde{k}$ Now what I want to do is integrate this momentum equation over the control volume and over time. Since it's 1-D I only consider a control volume around a single grid point P, the grid point East of point P is referred to as E and to the West is referred to as W. Let's say I want to integrate the term $\int_t^{t+\Delta{t}}\,\int_P^E\, \frac{1}{r^2}\frac{\partial}{\partial{r}}\left( r^2\overline{\rho}{\tilde{u}\tilde{u}}\right) \, dr \, dt$ If the term wasn't there this would have been straightforward, which is the case in the Cartesian coordinates. So one thing I tried to do was to multiply the whole equation by . But then I get terms like $\int_t^{t+\Delta{t}}\,\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr\,dt$ which again gives me the same problem. This may be different to what's normally done in Control Volume analysis. I'm trying to use the method of Patankar "Numerical Heat Transfer and Fluid Flow", 1980. Thanks. Last edited by lost.identity; April 16, 2010 at 11:19.

April 16, 2010, 13:41		#11
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Thanks, I think I sort of know what you're saying. Are you saying that when I multiply by , and integrate then terms such as $\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr= \Delta{V}\left [ (\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P \right ]$ is this true for 1-D case such as mine? and would $\Delta{V}=1$ ? or is $\Delta{V} = r_P^2\Delta{r}$ where is the radial value at the mid-point of the control volume? Last edited by lost.identity; April 16, 2010 at 15:42.

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April 14, 2010, 22:56		#2
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	use the control volume center coordiante to compute the 1/r^2.

April 15, 2010, 05:01		#3
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	My problem is actually with integrating the equation. I'm using Favre averaged version of the equation so I have the following convection term (1/r^2)d/dr(r^2rhouu) because u (the velocity) are functions of r I need to integrate this by parts. I tried multiplying the whole equation by r^2, but then this causes the same problem to other terms that doesn't have the (1/r^2) term. I've tried searching through the literature to find any papers that does finite volume method on spherical Navier-Stokes equation and haven't been able to find any that have done the same thing as what I'm trying to do.

April 15, 2010, 06:16		#4
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	multiply r2 at both sides

April 15, 2010, 07:48		#5
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Thanks for the reply. As I've already mentioned in my email, I've already tried that method. But the problem is when I multiply by r^2 on both sides I get terms like r^2dp/dr, which again has to be integrated by parts. But since p is generally a function of r this causes problems in the integration. In addition I have more complex terms that arises due to Favre averaging such as r^2 d/dr(<rhou''u''>), where <> refers to Favre averages. I use the Boussinesq approximation to model this but still I get terms such as ru, which again needs to be integrated by parts. Surely, there's another way to do this?

April 15, 2010, 18:59		#6
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	nope do not integrate by part. take r^2 is a constant

April 16, 2010, 05:19		#7
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	r^2 is not a constant, it arises by writing the Navier-Stokes equation in spherical form. r is one of the variable dimensions. So when I'm integrating with respect to r, r^2 can't remain a constant.

April 16, 2010, 08:57		#8
agd Senior Member Join Date: Jul 2009 Posts: 351 Rep Power: 18	I'm not sure I'm following what you are actually trying to do at this point. It sounds as if you are trying to integrate the full Navier-Stokes equations analytically. If you want to do a finite volume discretization, then the 1/r**2 term will be a part of what you discretize - in that case you will use the value at the cell center, as ztdep has suggested. Treat it just like any other term in the discretized equation.

April 16, 2010, 12:07		#10
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	r^2 tem represent the element volume in the sperical coordinate system Dv. as in cartesian coordinate, it is 1 dx dy dz.

April 16, 2010, 21:36		#12
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	yes, that is what i mean/ by the way , how do you input the formular?

April 17, 2010, 17:06		#13
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	THanks Just want to confirm, you mean $\Delta{V} = r_P^2\Delta{r}$ The site uses Latex formatting to type equations. If you click any of the equations you'd see the code.

April 18, 2010, 12:39		#14
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Actually shouldn't it be $\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr = \Delta{V} \left [\frac {\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \right ]_e= \Delta{V} \left [ \frac{(\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P}{\Delta{r}} \right ]$ where $\Delta{V} = r_e^2\Delta{r}$ , P and E refers to the left and right hand side faces control volume respectively, and e is the midpoint of the control volume. which would finally give $\int_P^E\,r^2\frac{\partial}{\partial{r}}(\overline{\rho{u''}{u''}}) \,dr = r_e^2 \left[ (\overline{\rho{u''}{u''}})_E-(\overline{\rho{u''}{u''}})_P \right ]$