initial condition in unsteady

hamid1 · February 22, 2011, 07:48

Hi!

Speaking for unsteady problem If u don’t know the initial condition, are the following approaches correct?

1. You may just guess the i.c. solving first in steady-state until well converge and then using it as i.c. for unsteady

OR

2. Using any arbitrary i.c. while it is solving unsteady and after some amount of time (or cycles if you have cyclic b.c.) we have reliable result –of course all in case of well converge-

Amir · February 22, 2011, 08:38

Hi,
can you solve 1 equation with two unknown variables!? what if you guess one of them? is the result reliable?!
of course not.

hamid1 · February 22, 2011, 11:28

Yes , I know my question is stupid… if y=f(x,t) then it is function of t as well as x, but something is vague for me when using a numerical approach CFD (like fluent), it is obvious that the initial condition they use in fluent is not a real i.c. because just for example the i.c. is basically the same for whole domain which is not correct in real physic or happens very rare, but as far as I know this Not correct i.c. will be vanished during the steps/time/solution, May be I am completely wrong:-D

February 22, 2011, 07:48	initial condition in unsteady	#1
hamid1 Senior Member hamid Join Date: Nov 2010 Posts: 185 Rep Power: 15	Hi! Speaking for unsteady problem If u don’t know the initial condition, are the following approaches correct? 1. You may just guess the i.c. solving first in steady-state until well converge and then using it as i.c. for unsteady OR 2. Using any arbitrary i.c. while it is solving unsteady and after some amount of time (or cycles if you have cyclic b.c.) we have reliable result –of course all in case of well converge-

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February 22, 2011, 08:38		#2
Amir Senior Member Amir Join Date: May 2009 Location: Montreal, QC Posts: 735 Blog Entries: 1 Rep Power: 22	Hi, can you solve 1 equation with two unknown variables!? what if you guess one of them? is the result reliable?! of course not.

February 22, 2011, 11:28		#3
hamid1 Senior Member hamid Join Date: Nov 2010 Posts: 185 Rep Power: 15	Yes , I know my question is stupid… if y=f(x,t) then it is function of t as well as x, but something is vague for me when using a numerical approach CFD (like fluent), it is obvious that the initial condition they use in fluent is not a real i.c. because just for example the i.c. is basically the same for whole domain which is not correct in real physic or happens very rare, but as far as I know this Not correct i.c. will be vanished during the steps/time/solution, May be I am completely wrong:-D