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April 28, 2016, 17:28 
How to fix / solve local instability

#1 
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Hector Redal
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Hi,
I would like to know how a local instability can be fixed. Is there any strategy when choosing the parameters for the solver? I am attaching an image of a local instability I am experiencing when simulating a flow past a circular cilinder. The flow is from left to right. Best regards, Hector. 

April 29, 2016, 03:13 

#2 
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Filippo Maria Denaro
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but running this solution for long time you get a numerical blowup of the solution? Apparently it seems that something in the BC.c can be wrong...
However, check for the CFL and viscous stability parameters in that local region 

April 29, 2016, 06:14 

#3 
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Hector Redal
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Hi Filippo,
The instability appears after the flow has been stablished and had reached the end of the domain (right). The boundary conditions I have using at the end of the domain are: du/dx = dv/dx = 0. The CFL is far below the critical CFL (by a factor of 2 or more). Best regards, hector. 

April 29, 2016, 07:50 

#4 
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Filippo Maria Denaro
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thus, this is not a numerical instability but a wiggle in the steady solution... check in which variables (those resolved in the code) you see such oscillations


April 29, 2016, 08:28 

#5  
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Arjun
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Quote:
Could you check if you have reversed flow at this point of problem. This information is important before any judgement be made because i know one way it could arise in case of reversed flow. 

April 29, 2016, 10:01 

#6 
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Hector Redal
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Hi Arjun,
After the instability appears, the flow reverses and start growing and growing until it becomes very large. The flow is supposed to be incompressible. There are two parameters that control the time marching scheme (relaxation factors) that can be varied. Theta1 for velocity which I set to 1.0 (It can be changed from 0.0 to 1.0) Theta2 for pressure which I set to 1.0 (It can be changed from 0.0 to 1.0) Best regards, Hector. 

April 29, 2016, 10:02 

#7  
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Hector Redal
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Quote:
Maybe I have misunderstood you in my reply. When the instability appears, there is not any steady state solution that can be reached. 

April 29, 2016, 10:27 

#8  
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Filippo Maria Denaro
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Quote:
Since your case is incompressible, I suggest to check the Div v = 0 constraint near the outflow 

April 29, 2016, 10:42 

#9  
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Arjun
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I think what he means is that wiggle appears but solver is able to recover from it.
Instability is some disturbance that keep growing until solver blows up the residuals. Quote:
So if you specify pressure Pbnd at boundary, in case of reversed flow it will be changed to Pbnd  1/2 density velocity sqr It is done to remove reversed flow and it generally helps. But sometimes what happens is that due to sudden change of boundary pressure from Pbnd to Pbnd  dynamic head, the pressure gradient at that cell sees sudden change. What it means is that sudden change in fluxes and other things in that cell. Sometimes solver becomes instable or produces wiggles due to this. I suspect what you see is this thing. 

April 29, 2016, 10:43 

#10 
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Arjun
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For incompressible flow pressure urf of 1 is too high, I would hesitate to try above 0.3 usually.
Only in pressure based coupled solver that i have i use p urf around 0.9 

April 29, 2016, 10:45 

#11  
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Hector Redal
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Quote:
Then I think that I am facing up an instability, since the solution blows up. I have double check the boundary condition (at the outflow), and it appears to be well posed: du/dx = 0 and dv/dx=0, which I understand this is the normal way of specifying the constraints at exit flow. On the other hand, I have been playing around with the time step (delta t), and after incresing the value of the delta t (from 0.001s to 0.005s) it seems it is working now. Why is this happening? I have always believed that reducing the time step means incresing the accuracy of the solution. This is someting that is puzzling me now. ??? 

April 29, 2016, 10:47 

#12  
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Filippo Maria Denaro
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Quote:
are you solving a Poisson pressure equation? check if the continuity constraint is satisfied by using your discretized BC.s. It is just sufficient check in a single time step what happen at the outflow 

April 29, 2016, 10:50 

#13  
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Hector Redal
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Quote:
By the way (only so as to frame the discussiion, in case it helps), the algorithm I am using is the Characteristics Based Split Algorithm which is a projection scheme for the Finite Element Method. 

April 29, 2016, 10:52 

#14  
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Arjun
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Quote:
What solver are you using btw. Decreasing time step may introduce decoupling of pressure and velocity as it weakens the Rhie and Chow dissipation. 

April 29, 2016, 17:29 

#15  
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Hector Redal
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Quote:
I do not use any kind of Rhie and Chow dissipation. Where can I find information about the Rhie and Chow dissipation. Do you know of any paper I can take a look at? 

April 29, 2016, 17:50 

#16 
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Filippo Maria Denaro
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your problem is in the outflow BC.s, the dissipation term is used in cellcentred colocation to avoid pressure decoupling...
Check the BC for the pressure problem 

April 30, 2016, 11:28 

#17  
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Hector Redal
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Quote:
I have checked the BC for the pressure and the velocity at the exit of the domain, and they appear to have been defined correctly. I have also checked the gradient of the velocity at the exit, according to your previous suggestion, and you can see in the attached images that they are well defined. I am attaching images for both gradients du/dx and dv/dx. At the exit it can be seen that they are both zero. By the way, I have increased a bit more the delta t of the resolution and it appears is working as well. Even the frequency of the vortex shedding approaches the value obtained by other references. 

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