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February 9, 2011, 17:57 
initila condition

#1 
Senior Member
hamid
Join Date: Nov 2010
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Hello Everybody
My question is does different initial condition in steady and non steady flow lead to different result? how can one initialize in the best way? thanks 

February 10, 2011, 04:40 

#2 
Senior Member

Hi hamid,
if you use enough iterations per time step in order to reach convergence criteria in both cases, your result will not be different. this is a valuable tip for steady cases in which convergency problems exist! 

February 11, 2011, 04:44 

#3 
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Join Date: Jun 2009
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Hi,
In steady cases initial conditions do not matter with regards to the final converged solution, however the closer your initial conditions are to the converged solution the lesser the number of iterations are required to reach the converged solution. On the other hand in unsteady you need to know the initial conditons and if you give the wrong initial conditions, the solution goes awry because in an unsteady case the every time step is based on the solution of the preceding time step. So if you give wrong initial conditions your entire solution is wrong. 

February 11, 2011, 19:01 

#4 
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hamid
Join Date: Nov 2010
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thank you for reply,
so plz just imagine that I have an unsteady problems, if first I solve it in steady mode (with arbitrary I.C.) until it converge and then shifting to Unsteady mode, then in this situation with any i.c. i will reach to the correct result in unsteady? 

February 12, 2011, 10:03 

#5 
Senior Member

Hamid,
if you solve steady case and then shift to unsteady solver, the results of converged steady solution will be your I.C. for unsteady ones! if you set another I.C. here by initializing or patching, your steady data will vanish! 

February 12, 2011, 17:02 

#6 
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hamid
Join Date: Nov 2010
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yeah, exactly.. i meant as you said kindly, i.e. with any i.c. (arbitrary), solving the problem first in steady mode and leading to a well converged result then using the result as i.c. for unsteady problem will lead to a correct solution...
but I did it for one of my problems as follow: there are two pipes attached to each other in the way that there are three openings, say opening 1,2 and 3, all three openings are supposed to be out let pressure, opening 1= 13300pa, opening2=13300 and opening 3(suction)= 100000 pa, for initialization first I made a mass balance between opening 3 and 2 and in other time I made a mass balance between opening 3 and opening 1&2, but the results are totally different ?? any clues or hint will diffidently help me, thanks 

February 13, 2011, 14:42 

#8 
Senior Member
hamid
Join Date: Nov 2010
Posts: 185
Rep Power: 9 
Hi Amir,
thanks for reply, then would u forward me ur email add I will send u the pics of b.c and IC by attachment 

February 14, 2011, 05:14 

#9 
Senior Member

Hi,
your pictures recieved. as I understood, non of your parameters vary with time e.g. p=p(t) or ,... are they? in these cases you just have a steady problem that converge to result from an initial guess that depend on schemes that you used. 

February 14, 2011, 06:11 

#10 
Senior Member
hamid
Join Date: Nov 2010
Posts: 185
Rep Power: 9 
thanks for reply, actually there are two phases (red and blue in which red has higher viscosity and density), by implementing the intake opening(3), the fluid moves and the idea is to absorb the red phase inside the opening 3, I think there is an unsteady problem, as u can see in one case the trend is red phase absorbing through opening 3 and in the other case (with other i.c.) the red phase is going to way out from another opening, plz kindly let me know what do u think then,


February 14, 2011, 12:17 

#11 
Senior Member

Hamid,
if you want to see final phase positions, it's steady problem. but if you want to trace phase changing from specific initial phase positions then u should use unsteady solver and using steady solution here is not reasonable. 

February 14, 2011, 12:51 

#12 
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hamid
Join Date: Nov 2010
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thanks, yes I want to see the movement of the second phase (red one) so it would be unsteady, but for i.c. could it influence my result, why I have different result for case 1 and case2 initialization? does solving in steady first and then using the result as new i.c. for unsteady solve my problem, i mean the steady solution works like a filter which whatever is my first i.c. , the coming out results is a good i.c. for unsteady,
thanks again, 

February 14, 2011, 16:33 

#13 
Senior Member

Ok, notice that I.C. in unsteady cases is a part of your problem it's not a first guess! you don't have permission to set it by chance. as you said, you want to trace second phase from t=0 to desired time, it will absolutely depend on your I.C..
it would be surprising if you got same results with different I.Cs. ! if you trace second phase for long time and your case doesn't have any special features to be oscillated for example, your final phase position may be independent of I.C. ,here as I said before you can use steady solvers. 

February 16, 2011, 06:27 

#14 
Senior Member
hamid
Join Date: Nov 2010
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But the i.c. is considered in all the domain in fluent, as far as I understand, so it would be kind of initial guess, is it right? then how can I find the exactly correct i.c.?
bests, hamid 

February 16, 2011, 09:36 

#15 
Senior Member

Right, I.C. is for whole domain but it's not a guess it's initial system condition e.g. at t=0. for unsteady problems I.C. should be given to you like boundary conditions. you can't find I.C. just like B.C.
you will get different results for different boundary condition just like I.C.. 

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