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September 18, 2004, 04:17 |
Cylindrical NS(FDM) at r=0
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#1 |
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Hi, I have a question about cylindrical NS equations. They have 1/r*(r*u) terms, which cannot be disretized at r=0. I am using staggerd 2D axisymmetric cylindrical cordination and would like to give value at r=0. I have tried
--extrapolation --1/r*(r*u) = (u - (-u))/dr, (axisymmetric) The program runs, but is it OK? Have you got any suggestion? |
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September 18, 2004, 06:17 |
Re: Cylindrical NS(FDM) at r=0
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#2 |
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In the axisymmetric cylindrical NS equations, the radial velocity u is zero at r=0.
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September 18, 2004, 09:54 |
Re: Cylindrical NS(FDM) at r=0
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#3 |
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L'hospital rule: u/r = du /dr -> (u(i+1)-u(i))/dr +O(dr^2)
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September 20, 2004, 02:04 |
Re: Cylindrical NS(FDM) at r=0
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#4 |
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Hi
goto : http://ctr.stanford.edu/ then : Publications Then : annual research brief 2000 Search fo Lele, The paper is A new method for accurate treatment of flow equations in cylindrical coordinates using series expansions. G. S. Constantinescu and S. K. Lele Regards Apurva |
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September 25, 2004, 05:02 |
Successive Quenstion
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#5 |
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Thank you for your kind reply. My program is almost complete. But I have one more problem.
I know Vz/r = d(Vz)/dr = 0 but what about Vr/r = d(Vr)/r = ?? Which is appropriate? Vr/r = d(Vr)/r = (Vr(1) - 0)/dr or Vr/r = d(Vr)/r = 0 |
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