viscosity term in discetrization momentum equation

hans-186 · March 31, 2013, 07:38

Hi all,

I have a question regarding the discretization of the momentum equation for use in a simple solver.
I'm having trouble in understanding where the viscosity term goes. I'm currently reading Versteeg 2007 , An introduction to CFD, and can't figure it out.

At some point the momentum equation, including the body surface forces caused by viscosity is discretisized to:

$a _{i,J}*u _{i,J}=\sum a_{nb}*u_{nb}+ \frac{ P_{I,J}-P_{I-1,J} }{ \delta x }*\Delta V+S*\Delta V$

As I understand it, the viscosity term is not present here, coefficients $a_{n,n}$ only have a velocity and density dependence. I can't seem to locate a viscosity in the source term,S ,either.

If find that on the web it is often pointed out that the discretization is similar to that of the general transport equation where the viscosity term can be handled equal to the diffusion term and the velocity term to the property term. In this case the diffusion term is preserved in the discretization, why than not (or at least lost to me) in the momentum equation?

Can anyone explain what i'm missing or point me in a direction where i can find some answers?I would be very happy to find the complete derivation of the discretization of the momentum equation so i can see what's happening step by step.

Kind regards,

FMDenaro · March 31, 2013, 07:51

Quote:

Originally Posted by hans-186

Hi all,

I have a question regarding the discretization of the momentum equation for use in a simple solver.
I'm having trouble in understanding where the viscosity term goes. I'm currently reading Versteeg 2007 , An introduction to CFD, and can't figure it out.

At some point the momentum equation, including the body surface forces caused by viscosity is discretisized to:

$a _{i,J}*u _{i,J}=\sum a_{nb}*u_{nb}+ \frac{ P_{I,J}-P_{I-1,J} }{ \delta x }*\Delta V+S*\Delta V$

As I understand it, the viscosity term is not present here, coefficients $a_{n,n}$ only have a velocity and density dependence. I can't seem to locate a viscosity in the source term,S ,either.

If find that on the web it is often pointed out that the discretization is similar to that of the general transport equation where the viscosity term can be handled equal to the diffusion term and the velocity term to the property term. In this case the diffusion term is preserved in the discretization, why than not (or at least lost to me) in the momentum equation?

Can anyone explain what i'm missing or point me in a direction where i can find some answers?I would be very happy to find the complete derivation of the discretization of the momentum equation so i can see what's happening step by step.

Kind regards,

in § 6.3 it is explicitly stated that the coefficients contain combination of convective and diffusive terms..

hans-186 · March 31, 2013, 07:56

FMDenaro,

Thanks for your reply!

I've seen this in chapter 6.3 .But I'm a bit confused in how to interpret this, the viscosity and diffusive terms are the same?
I presume this is the diffusion of momentum from one cell to the other? Hoe to couple this momentum diffusion term to viscosity then?

thegodfather · April 1, 2013, 13:17

Viscosity by definition is the momentum diffusivity.. The rate at which momentum of one layer of the fluid is diffused to other layers is decided by the viscosity..

hans-186 · April 1, 2013, 14:34

Yeah, I'm back on track

. Got somewhat lost in the maze I guess.
Thanks for your replies.

March 31, 2013, 07:38	viscosity term in discetrization momentum equation	#1
hans-186 New Member hans Join Date: Sep 2012 Posts: 7 Rep Power: 14	Hi all, I have a question regarding the discretization of the momentum equation for use in a simple solver. I'm having trouble in understanding where the viscosity term goes. I'm currently reading Versteeg 2007 , An introduction to CFD, and can't figure it out. At some point the momentum equation, including the body surface forces caused by viscosity is discretisized to: $a _{i,J}u _{i,J}=\sum a_{nb}u_{nb}+ \frac{ P_{I,J}-P_{I-1,J} }{ \delta x }\Delta V+S\Delta V$ As I understand it, the viscosity term is not present here, coefficients $a_{n,n}$ only have a velocity and density dependence. I can't seem to locate a viscosity in the source term,S ,either. If find that on the web it is often pointed out that the discretization is similar to that of the general transport equation where the viscosity term can be handled equal to the diffusion term and the velocity term to the property term. In this case the diffusion term is preserved in the discretization, why than not (or at least lost to me) in the momentum equation? Can anyone explain what i'm missing or point me in a direction where i can find some answers?I would be very happy to find the complete derivation of the discretization of the momentum equation so i can see what's happening step by step. Kind regards,

March 31, 2013, 07:56		#3
hans-186 New Member hans Join Date: Sep 2012 Posts: 7 Rep Power: 14	FMDenaro, Thanks for your reply! I've seen this in chapter 6.3 .But I'm a bit confused in how to interpret this, the viscosity and diffusive terms are the same? I presume this is the diffusion of momentum from one cell to the other? Hoe to couple this momentum diffusion term to viscosity then? Last edited by hans-186; March 31, 2013 at 08:25.

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April 1, 2013, 13:17		#4
thegodfather New Member Aniket Sachdeva Join Date: Mar 2012 Posts: 22 Rep Power: 14	Viscosity by definition is the momentum diffusivity.. The rate at which momentum of one layer of the fluid is diffused to other layers is decided by the viscosity..

April 1, 2013, 14:34		#5
hans-186 New Member hans Join Date: Sep 2012 Posts: 7 Rep Power: 14	Yeah, I'm back on track . Got somewhat lost in the maze I guess. Thanks for your replies.