
[Sponsors] 
April 23, 2018, 09:43 
Couette Flow

#1 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Hello to everyone,
I am writing SIMPLE code for couette Laminar flow with the existence of pressure. I have read similar topics for this problem on this forum but it is not clear yet. My question is about: 1) How can I apply the given pressure gradient in the SIMPLE algorithm? Maybe in the provisional velocities equations u*,v* can I replace the term p(i+1,j)p(i,j) with the given pressure gradient? 2) Can you suggest me the correct boundary conditions at inlet, outlet, walls for the horizontal velocity, vertical velocity and pressure? Thank you. Last edited by IbizaFR; April 23, 2018 at 10:58. 

April 23, 2018, 11:46 

#2  
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Quote:
You have to employ periodic bcs at inlet/outlet and add the pressure gradient as body force. In practice, you have to add to your momentum equation a term like dpdx*vol, where vol is the cell volume. For the pressure equation, set periodic bcs for inlet/outlet, ehile simple zero gradient for solid walls. 

April 23, 2018, 12:31 

#3 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Thank you very much.
So for the xmomentum I add this term dp/dx*vol. In the ymomentum? what are periodic bcs? Can you state about the velocities in boundaries? 

April 23, 2018, 12:40 

#4 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Your source term is naturally a vector. Since you are simulating a couette flow only xmomentum is non zero. Thus let's say that you want to impose a gradient of 0.25, the add at each cell 0.25*volume. The sign will determine the flow direction.
For the moving wall you can set a standard (1 0 0) fixed velocity while (0 0 0) for the other. Periodic bc means that you inlet velocity will take the outlet velocity and Pressure as boundary condition. Immagine to have a rigid translation of your channel in order to produce an infinite channel. Those conditions are used to simulate something repeting infinitly such as couette or channel flows. In practice you must consider the bundary faces as internal faces. 

April 23, 2018, 12:48 

#5 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
I understand.
So in the xmomentum I add lets say 0.25*volume and in the ymomentum nothing. For the velocities at the moving wall I have u=1 (dimensionless), v=0 and at the stationary wall u=v=0. For the inlet what is the relationship that connects with outlet velocity and pressure? Thank you. 

April 23, 2018, 13:05 

#6 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Yes, everithing is understood as non dimensional. In order to create a couple of periodic boindaries you have to adjust a bit the data structure of you mesh. For examole yuo can create a couple of "ghost" columns cointaing the values of velocity and pressure at I=1 and I=NI. If your code is not parallel you can choose other simple options, but you have to make sure that the outlet cells know the inlet cells values and vice versa. Just to give you an example, if you want to calculate the face velocity at outlet by linear interpolation it would be equalt to uf=(u1+uN)/2 where u1 and uN are the velocity values at the cell centers for the first and the last sets of cells along ydirection in your channel. The same applies for other physical quantities. Your coeffs will become the same as the internal face case. As you can see you should "cheat" your code in order to mimic an infinite channel.


April 23, 2018, 13:10 

#7 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Thank you AliE,
Before continuing in order to make it crystal clear is it right for the walls to take θp/θn=0 where n is the normal direction? 

April 23, 2018, 13:21 

#8 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Sure. Pressure is periodic at inlet/oultlet and zero grad at solid walls. Also look at this paper which is very usefull to you:
Implementation of boundary conditions in the finitevolume pressurebased method—Part I: Segregated solvers. By Darwish et al. 

April 23, 2018, 13:31 

#9 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Thank you.
So for the inlet and outlet I have to relate the velocity and the pressure in these two positions. I am right? 

April 23, 2018, 14:05 

#10 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Yes think the inlet as the neoghbour of the outlet.


April 24, 2018, 04:20 

#11 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Thank you AliE,
Lets take the simple Couette flow where dp/dx=0. So I have to put zero all the terms regarding the pressure gradient in the x and y momentum provisional velocities equations. Is it correct this? In this case do I have in my code pressure correction equation since there is not any pressure? 

April 24, 2018, 04:41 

#12 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Sure. You have to see the pressure eq as sonething that enforce and guarantee mass consevation (continuity eq). In you simple case however the pressure field will ve constan.


April 24, 2018, 04:43 

#13 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Do I solve pressure correction equation in this case?


April 24, 2018, 04:44 

#14 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Yes solve it in any case.


April 24, 2018, 04:47 

#15 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
I solve it and the pressure correction calculates the velocity correction?


April 24, 2018, 04:49 

#16 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
Yes this is thr standard alghoritm. If the pressure correction will be constant than its gradient will be zero e there will be no correction. But this is a parcicular case and in any case if you have problem with bcs your code will diverge highlithing the error.


April 24, 2018, 04:53 

#17 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
Thank you,
So I solve pressure correction equation then calculate velocity correction and then I enforce constant presssure everywhere e.g. p(1:nx,1:ny,n+1)=1 where nx, ny the x,y nodes and n the itelation level. Is it correct? 

April 24, 2018, 05:02 

#18 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
No. Solve momentum, solve pressure correction, correct velocity field and pressure field (p=p+p'). The pressure is a result of the computation.


April 24, 2018, 05:08 

#19 
Member
E
Join Date: Mar 2018
Posts: 74
Rep Power: 5 
How will I enforce constant pressure field everywhere since I have only simple couette flow?


April 24, 2018, 05:32 

#20 
Senior Member
Join Date: Dec 2017
Posts: 153
Rep Power: 5 
You are misunderstanding a bit what you are going to do.
You have not to impose a constant pressure. The pressure is a reuslt of your calculation and it is expected to be constant in your case (and only in your particular case). So simply follow the steps of alghoritm. If you have a pressure gradient, yhen add it as body force. 

Thread Tools  Search this Thread 
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Issues on the simulation of highspeed compressible flow within turbomachinery  dowlee  OpenFOAM Running, Solving & CFD  9  February 5, 2021 10:15 
FLuent simulation of taylor couette flow of concentric cylinder geometry.  rshbhb  FLUENT  53  November 5, 2014 19:07 
Couette flow in fluent  dreamz  Main CFD Forum  19  July 18, 2014 11:13 
Verification: Couette Flow  AndrewLBloxom  STARCCM+  0  March 28, 2014 09:58 
2D Couette/ Poisuelle Flow  CFD Guy  FLUENT  1  April 30, 2011 09:48 