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URGENT ! Need help on Axisymmetric Flow !

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Old   November 20, 2001, 11:33
Default URGENT ! Need help on Axisymmetric Flow !
  #1
Suman Kumar
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Hi,

I need urgent help! Can anybody help me understanding the finite-volume method for axisymmetric flow? For theory and implementation I am using " Computational Methods for Fluid Dynamics- 2nd Edition." by Joel H. Ferziger & M. Peric. and accompanying CAFFA code. I have two points regarding this :

1) On pages 244-246 it explains that general Cartesian code can be adopted for axisymmetric flow if some additional terms are added. Can you please help me understand why it is needed to add the radial component of pressure forces on the front and back surfaces?(page 245) I am also not very clear how this extra term is derived in here. There is a very sort explanation of this fact in here. Is it possible to give some references which treats this in greater details?

2)By studying CAFFA code it appears that gradient vector components at CV-centre for U, V and P all use above mentioned correction terms for axisymmetric flow. Subroutine for calculating gradient is

GRADFI(K,FI,DFX,DFY)

In this subroutine correction for axisymmetric flow is provided and this subroutine is called by another subroutine

CALCUV(K)

This calculates the velocities component. My question here is -- why is it required to add correction term(1)for all the gradients ? I will greatly appreciate any help in terms of explanation or reference.

Thank you very much !

Suman Kumar.
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Old   November 20, 2001, 15:51
Default Re: URGENT ! Need help on Axisymmetric Flow !
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Patrick Godon
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In an axisymmetric onfiguration, you can use sperical coordinates or cylindrical coordinates. Now I guess that your model is making use of cylindrical coordinates, Therefore you do have r (radial coordinate), theta (angular coordinate) and z (vertical coordinate). Now if the flow is axisymmetric you dont need to bother about theta, but just about r and z. You are consequently left with the plane (r,z). The z axis, is the r=0 axis. The axis r and z are normal to each other and you can (if you wish) use (x,y) instead of (r,z). However, when you are in the small r (x), you have actually a cylinder with a small radius, while when you are in the large r (x) you actually have a cylinder with a large radius. Therefore the surface of these cylinder is increasing with radius, and it increases like r itself. Therefore when you consider the conservation equations of hydrodynamics, where you actually look at the fluxes of quantities (momentum, mass), the surfaces through which these fluxes are advancing in the radial direction is increasing with r, and therefore one has to correct them accordingly (since the flux - say of mass - is the mass per unit time passing through a unit of surface). When one looks at the pressure force, for sure the pressure force is equal to the pressure times the surface, and when the surface you are considering are not equal (but are a function of r), one needs to take that into account. So, when one writes that you can use cartesian coordinates with additional terms, it really means that one uses (x,y) to represent (r,z). Then (x,y) is really not 'cartesian' properly speaking, but just letters!

FOr the same reason, the gradient needs to be corrected.

What you want to do, is, to get the equations written in cylindrical coordinates, then get rid of all the derivatives in the angular direction. Then replace r by x and z by y, and you'll get what you want. This is an easy exercise, just take book with the Navier Stokes equations written in cylindrical coordinates.

I hope this helps.
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