2D and 2D axis difference

entropies · January 27, 2012, 00:07

I have a euler-solver code, structured grid. It works for 2D. I want to change it to 2d axisymmetric. I've changed the associated volume and face area and source term coming along with axisymmetric.

But there's still problem in it. Is there anything else I should do?

Thank you.

Martin Hegedus · January 27, 2012, 01:59

Source term?

You'll need to change the fluxes (RHS) in the j (theta direction) along with the required implicit stuff on the LHS (if you are doing implicit).

For the RHS you have a flux term (which does not exist in 2D) in the theta direction along the lines of p*A(j+1)-p*A(j-1) but A(j+1)-A(j-1) can be reformulated as -(A(i+1)-A(i-1)+A(k+1)-A(k-1)). Or something along those lines. That was for central differencing.

That, along with changes in area and volume, should be enough. I believe...

Martin Hegedus · January 27, 2012, 02:02

Oh, one more very important thing. Unlike 2D, your CFL number MUST take into account the theta direction. I usually use a 5 degree wedge to calculate CFL.

January 27, 2012, 00:07	2D and 2D axis difference	#1
entropies New Member Qi Guan Join Date: Aug 2011 Posts: 27 Rep Power: 14	I have a euler-solver code, structured grid. It works for 2D. I want to change it to 2d axisymmetric. I've changed the associated volume and face area and source term coming along with axisymmetric. But there's still problem in it. Is there anything else I should do? Thank you.

January 27, 2012, 01:59		#2
Martin Hegedus Senior Member Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19	Source term? You'll need to change the fluxes (RHS) in the j (theta direction) along with the required implicit stuff on the LHS (if you are doing implicit). For the RHS you have a flux term (which does not exist in 2D) in the theta direction along the lines of pA(j+1)-pA(j-1) but A(j+1)-A(j-1) can be reformulated as -(A(i+1)-A(i-1)+A(k+1)-A(k-1)). Or something along those lines. That was for central differencing. That, along with changes in area and volume, should be enough. I believe...

January 27, 2012, 02:02		#3
Martin Hegedus Senior Member Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19	Oh, one more very important thing. Unlike 2D, your CFL number MUST take into account the theta direction. I usually use a 5 degree wedge to calculate CFL.