# low speed compressible flow

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 October 27, 2016, 07:08 low speed compressible flow #1 New Member   Abolfazl Join Date: Oct 2016 Posts: 28 Rep Power: 9 Hi everyone. I am trying to simulate compressible viscous flow around a circular cylinder by implementing VAN-LEER scheme. for supersonic flow the results are satisfactory but for low mach number (mach=0.3 and Reynolds=300) I can't simulate the vortex shedding. I only get two stationary small vortexes. can anybody suggest me what should I do!?? thanks.

October 27, 2016, 07:24
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by Abolfazl_cfd Hi everyone. I am trying to simulate compressible viscous flow around a circular cylinder by implementing VAN-LEER scheme. for supersonic flow the results are satisfactory but for low mach number (mach=0.3 and Reynolds=300) I can't simulate the vortex shedding. I only get two stationary small vortexes. can anybody suggest me what should I do!?? thanks.
two steady symmetrical vortical structures indicate that you are resolving an actual low Reynolds number.... Try a grid refinement study.

October 27, 2016, 08:22
#3
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Abolfazl
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Quote:
 Originally Posted by FMDenaro two steady symmetrical vortical structures indicate that you are resolving an actual low Reynolds number.... Try a grid refinement study.
but my mesh is fine enough. the problem is as i increase the mach number the length of steady symmetrical vortexes increase and as i decrease the mach the length decrease.
I don't think this is related to mesh.

 October 27, 2016, 08:26 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,813 Rep Power: 73 Assuming the speed of sound is the same, you are changing both the Mach and the Reynolds number. If you decrease it and your scheme has numerical viscosity, your actual Re number can be as low as to produce a laminar steady solution. Ensure that at low Re you have no turbulence model active in your code.

October 27, 2016, 08:35
#5
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Abolfazl
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Quote:
 Originally Posted by FMDenaro Assuming the speed of sound is the same, you are changing both the Mach and the Reynolds number. If you decrease it and your scheme has numerical viscosity, your actual Re number can be as low as to produce a laminar steady solution. Ensure that at low Re you have no turbulence model active in your code.
thanks again. but the code is non-dimensionalized and I just set reynolds and mach number separately. so the speed of sound may not be constant at different mach numbers.
my numerical method is Van-Leer and it is first order upwind. and i don't have turbulence model.

 October 27, 2016, 08:51 #6 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,813 Rep Power: 73 the solution with first order upwind is full of artificial viscosity...check your grid is so fine to get the cell Reynolds number =O(1) everywhere

October 27, 2016, 09:24
#7
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Abolfazl
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Quote:
 Originally Posted by FMDenaro the solution with first order upwind is full of artificial viscosity...check your grid is so fine to get the cell Reynolds number =O(1) everywhere
thanks again.
I have checked with really fine grid and I'm almost certain the grid is fine enough.
but don't you think that Van Leer doesn't work properly for low mach numbers or for viscous flows!?

 October 27, 2016, 17:29 #8 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 Can't say if Van Leer will work or not, but, since your original post reads as if you may have written the code, did you divide the Reynolds number by the Mach number? For a non-dimensionalized compressible solver the Reynolds number is, in general, calculated using the speed of sound rather than the velocity therefore the inputed Reynolds number (which in general is calculated using velocity) is divided by the Mach number internal to the code.

October 30, 2016, 04:18
#9
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Abolfazl
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Quote:
 Originally Posted by Martin Hegedus Can't say if Van Leer will work or not, but, since your original post reads as if you may have written the code, did you divide the Reynolds number by the Mach number? For a non-dimensionalized compressible solver the Reynolds number is, in general, calculated using the speed of sound rather than the velocity therefore the inputed Reynolds number (which in general is calculated using velocity) is divided by the Mach number internal to the code.
Dear Martin.
No, I didn't divide the Reynolds number with Mach number. I just set Mach and Reynolds number. and the Reynolds number is base on free stream velocity.

 October 30, 2016, 09:47 #10 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 Given your reply, I'm not sure my post was understood correctly. In general, the Reynolds number that goes into compressible solvers is based on the freestream velocity. So your post reads as if your input is correct. However, internal to a compressible code the Reynolds number will be modified to be based on the speed of sound by dividing by the Mach number. So, if you have written your own code, make sure your equations are non-dimensionalized consistently.