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 F August 6, 2004 03:24

Confused

Hello,

A couple of questions that I cannot find answers to in several text-books.

1. What is the physical meaning of the term (u/r) that appears in cylindrical coordinates, continuity eqn.?

2. What use are cylindrical coordinates if my problem is axisymmetric, using only r and z directions? (the cartesian coordinates does not show the extra term so I'm a bit confused here.)

3. If question 2 has anything to do with the control volume approach used in FVM, then does it matter if I would use FDM instead?

I have not attended any fluid mechanic courses so please be gentle with your answers if this is all too simple. =)

/F

 Daniel August 6, 2004 08:03

Re: Confused

They are just some personal ideas and may be wrong:

Your N-S equations are differential forms of conservation laws, thus, when you use FDM in cylindrical c., also in the case of axisymmetry, you solve the equation in a plane (meridian plane), but this plane has to be considerd a limit to zero of finite 3d region; so your meridian plane becomes equivalent to a cartesian plane if you add some "terms" whose meanigs can be derived by the general conservation laws. If you use FVM, you directly discretize the conservation law, so you don't need the additional term, (BE CAREFULL, THE BEST THING IS TO SOLVE IN FVM IN THE MERIDIAN PLANE ONLY, IN THIS CASE YOU NEEDS THE ADDITIONAL TERMS)

FOr example: u/r+du/dr=1/r*d(ur)/dr (1) the right hand side is the flux of u through an elementary region , let' s say:

R1<r<R2

TETA1<teta<TETA2 You can easlily check by using the Gauss theorem. However, in (1) teta doesn't appear, but the additional term u/r accounts for its "existence". in conclusion, my sugestion is to not confuse cartesian 2d planes with meridian planes of cylindrical c. system.

 Daniel August 6, 2004 08:12

Re: Confused

I continue the previous answer, I was saying u/r +du/dr=1/rd(ur)/dr (1) The right hand side of (1) is the flux of u through an elementary region, say R1 < R < R2 Teta1<Teta<Teta2 (you can check by Gauss theorem) In (1) the variable teta doens't appear, however, the additional term u/r "accounts" for its existence. In conclution, a cartesian 2d plane and a meridian cylindric plane, ARE NOT the same thing, bacause they are the limit to zero of two different entities.

However, your question is not so easy! Bye

 Rami August 8, 2004 02:53

Re: Confused

Q1. What is the physical meaning of the term (u/r) that appears in cylindrical coordinates, continuity eqn.?

A1: Assumming u is the circumferential velocity (i.e., in theta direction), u/r is the circumferential strain-rate.

Q2. What use are cylindrical coordinates if my problem is axisymmetric, using only r and z directions? (the cartesian coordinates does not show the extra term so I'm a bit confused here.)

A2: If you solve an axisymmetric problem in Cartesian coordinates, you still need to consider 3 velocity components and to solve 3 momentum equations. It is true the equations are a bit simpler in this case. On the other hand, in cylindrical coordinates, you need to consider only 2 velocity components and 2 momentum eqautions if the circumferential velocity is 0 and the geometry is axisymmetric, but some extra terms appear in the equations in this system.

Q3. If question 2 has anything to do with the control volume approach used in FVM, then does it matter if I would use FDM instead?

A3: You may use either FDM or FVM, regardless of the coordinate system you choose. These are different numerical methods. FDM discretize the conservation equations by direct application of Taylor series ("strong formulation"), while the FVM uses the integral control volume conservasion ("weak formulation"). Each has its pros and cons, but the general trend favours FVM for CFD.

I hope this helps.

 F August 8, 2004 20:32

Re: Confused

Q1. U is the radial velocity.

Q2. Here I think you hit the core of my confusion. I don't see why cartesian coordinates means 3 directions if the problem is 2 dimensional. If a control volume approach is used then there is a volume to be considered (although thin), but if I use FDM, then I should be able to express the governing eqns. in only 2 directions, right? If you dispose of the azimuthal velocity and only consider radial and axial then isn't this true? (a cylinder watched from the side, i.e. rectangle)

Regards

 Rami August 9, 2004 02:16

Re: Confused

Q1. U is the radial velocity.

A1. The u/r term in the continuity eq. is merely part of the radial mass flux 1/r*d/dr(r*rho*u) . I am not sure if it has a physical meaning by itself.

Q2. Here I think you hit the core of my confusion. I don't see why cartesian coordinates means 3 directions if the problem is 2 dimensional. If a control volume approach is used then there is a volume to be considered (although thin), but if I use FDM, then I should be able to express the governing eqns. in only 2 directions, right? If you dispose of the azimuthal velocity and only consider radial and axial then isn't this true? (a cylinder watched from the side, i.e. rectangle)

A2. When you deal with axisymmetric geometry and flow, the symmetric part you should consider is a wedge-like sector, not a rectangular box. Although it may be thin, the non-parallel faces lead to significant terms, and also the symmetry BC are prescibed on them rather than on faces normal to one of the Cartesian coordinates. Please refer to any continuum mechanics (solid or fluid) textbook.

I hope this clarifies your confusion.

 F August 9, 2004 05:32

Re: Confused

Thanks again Rami.

Q1. Ok =)

Q2. To sum up; axisymmetry does not reduce a problem to 2d even if it looks that way, if you consider an azimuthal wedge of zero.

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