[mikethe1wheelnut]

Greetings all :-)


First, on this subject: Guide: How to ask a question on the forums, if I followed those guidelines to the letter, I would likely never post anything at all because I would never be desperate enough for the answer. I don't -need- the answer, it would just be nice to have, and maybe there is a reference somewhere I've missed.


My question is closely related to this other one, in this same part of the forum, concerns exactly the same pages of the same book: 2D axisymmetric Navier-stokes equation


My focus is instead the source-term. I have not succeeded, yet, in comprehending how switching to cylindrical coordinates introduces an extra component to the source term.



I have searched for answers to this online, but while I have found several derivations of navier stokes in cylindrical, none of them present their answers in the same format as this text.


The next thing I would try, if I really needed to know, would be to attempt to re-derive the equation, with the new symmetry, following material covered earlier in the book. I could also consult derivations of the equations presented as differential equations in a few places, and attempt to convert those to integral equations. From that, it might become obvious where the source term has come from..



Enjoy your coding!

[LuckyTran]

There is no source term in the Navier-Stokes equation so there is nothing to derive.

[Hamidzoka]

Originally Posted by mikethe1wheelnut

Greetings all :-)


First, on this subject: Guide: How to ask a question on the forums, if I followed those guidelines to the letter, I would likely never post anything at all because I would never be desperate enough for the answer. I don't -need- the answer, it would just be nice to have, and maybe there is a reference somewhere I've missed.


My question is closely related to this other one, in this same part of the forum, concerns exactly the same pages of the same book: 2D axisymmetric Navier-stokes equation


My focus is instead the source-term. I have not succeeded, yet, in comprehending how switching to cylindrical coordinates introduces an extra component to the source term.



I have searched for answers to this online, but while I have found several derivations of navier stokes in cylindrical, none of them present their answers in the same format as this text.


The next thing I would try, if I really needed to know, would be to attempt to re-derive the equation, with the new symmetry, following material covered earlier in the book. I could also consult derivations of the equations presented as differential equations in a few places, and attempt to convert those to integral equations. From that, it might become obvious where the source term has come from..



Enjoy your coding!

Hi,
If I am right, you are talking about terms like "viscosity*Ur*Ur/r**2". These are not source terms in its classical defintion. If you see the Laplacian and divergence definitions in cylindrical reference frame, you will find some coefficiencts like r and 1/r which are not the case in cartesian ones. This terms, when derivated, generate some terms that looks like source terms.

Best

[mikethe1wheelnut]

Originally Posted by Hamidzoka

Hi,
If I am right, you are talking about terms like "viscosity*Ur*Ur/r**2". These are not source terms in its classical defintion. If you see the Laplacian and divergence definitions in cylindrical reference frame, you will find some coefficiencts like r and 1/r which are not the case in cartesian ones. This terms, when derivated, generate some terms that looks like source terms.

Best

I am referring to the second attached thumb-nail image here: 2D axisymmetric Navier-stokes equation, which, when I click on it expands into a view of the text-book in question. There, you will find equation A.61, which is referred to again at the end of page 449, claiming that the conversion to cylindrical effectively introduces an extra source term, even if there wasn't one there originally. this source-term could be written as [0,0,(p-tau_theta_theta)/r,0], so only applicable to the radial component..

[FMDenaro]

Originally Posted by mikethe1wheelnut

I am referring to the second attached thumb-nail image here: 2D axisymmetric Navier-stokes equation, which, when I click on it expands into a view of the text-book in question. There, you will find equation A.61, which is referred to again at the end of page 449, claiming that the conversion to cylindrical effectively introduces an extra source term, even if there wasn't one there originally. this source-term could be written as [0,0,(p-tau_theta_theta)/r,0], so only applicable to the radial component..

Unfortunately, the terminology used by Blazek is misleading.

First, the general integral form of the momentum equation has only the surface integral of the normal component of the fluxes. There is no volume integral of a source Q (even in case of gravity we could introduce a potential and then convert in the surface integral).
The textbook denoted as "source" a term that appears only in the specific reference system, it has not physical meaning of source.
I supposed Blazek just organized that for the specific case of cylindrical system and the word "source" has not any physical meaning but only collect that terms outside the surface integral in the initial equation.

[mikethe1wheelnut]

Originally Posted by FMDenaro

Unfortunately, the terminology used by Blazek is misleading.

First, the general integral form of the momentum equation has only the surface integral of the normal component of the fluxes. There is no volume integral of a source Q (even in case of gravity we could introduce a potential and then convert in the surface integral).
The textbook denoted as "source" a term that appears only in the specific reference system, it has not physical meaning of source.
I supposed Blazek just organized that for the specific case of cylindrical system and the word "source" has not any physical meaning but only collect that terms outside the surface integral in the initial equation.

..this basically matches my understanding. Blazek has re-organized the terms to try to clarify what has actually changed, leading to -other- confusions. So I would have to, or I will have to, go through the derivation myself, possibly following online sources I've collected, to see for myself where this term actually comes from. Thanks!

[FMDenaro]

Originally Posted by mikethe1wheelnut

..this basically matches my understanding. Blazek has re-organized the terms to try to clarify what has actually changed, leading to -other- confusions. So I would have to, or I will have to, go through the derivation myself, possibly following online sources I've collected, to see for myself where this term actually comes from. Thanks!

My suggestion is to start from the differential divergence form of the momentum equation, then simply express the nabla operator and vectors in the specific reference system and apply the rule of inner product and derivation of unit vector.

Then you can analyse the integral form.

[LuckyTran]

Just to further clarify, the "source" term that you are specifically asking about is just the centripetal force. It already has a very clear and intuitive meaning to it.

[mikethe1wheelnut]

Originally Posted by LuckyTran

Just to further clarify, the "source" term that you are specifically asking about is just the centripetal force. It already has a very clear and intuitive meaning to it.

That -does- make a lot of sense. Particularly since I've gone back and looked at the ordinary source term again, and wrapped my head, more-or-less, around gravity itself being a source term. Chemical reactions generating heat.. those are obvious.. :-)

[FMDenaro]

Originally Posted by mikethe1wheelnut

That -does- make a lot of sense. Particularly since I've gone back and looked at the ordinary source term again, and wrapped my head, more-or-less, around gravity itself being a source term. Chemical reactions generating heat.. those are obvious.. :-)

Even gravity, under incompressible flow assumption, is not a source term since can be written as surface integral exactly as the pressure.