conditions for unsteady problems

dvdromnu · November 7, 2009, 02:42

Hi,
I am solving unsteady state convective diffusion scalar equation using finite volume method(explicitly).
My problem is:
For steady state problem the solution converges at some time step and we get results. That means steady state problem is independent of initial conditions.
For unsteady state problem I am printing the values of scalar at a particular time step.
So My question is, for unsteady problem what initial conditions I have to take..?
means if I run unsteady problem for long time as in case of steady problem I am getting the same result of Steady state problem.

So, how I have to solve unsteady problem. What initial conditions I should take. What is the difference between unsteady state and steady state solutions

alberto · November 7, 2009, 15:02

If the problem admits a steady state solution, the unsteady solution procedure should lead to the same solution, if it is run long enough.

dvdromnu · November 8, 2009, 01:45

Thank You................

alberto · November 8, 2009, 02:52

Quote:

Originally Posted by dvdromnu

Thank You................

I'm sorry for the very vague answer. If you want more details, you should provide some more information about your problem.

Best,

dvdromnu · November 9, 2009, 01:49

Actually I want to know whether there is any difference in solution procedure of unsteady state and steady state scalar convective diffusion equation.

Also I want to know if we solve scalar equation with diffusion term only(without convection term) we will get one solution. Now, if we solve the same including convection term, we get another solution. So, adding convection term changes the solution profile( with only diffusion term) or it shifts the solution profile..?

dmoroian · November 9, 2009, 03:15

Hello .... dvdromnu (actually it would be nice to have a name here),
If the equations you're solving admit a solution, and if that solution is unique, then it doesn't matter what initial conditions you start with (as long as they are physical or representative for your equations).
If you're interested in the steady state solution, then again it doesn't matter what approach you use (steady/unsteady).
If the convection/diffusion equation has the same steady state solution as the diffusion only equation, then you will not see any "profile shift".
The only difference would be in the choice of the solver, and the amount of computational time requested (shorter for steady state and longer for unsteady).

I hope this was useful,
Dragos

November 7, 2009, 02:42	conditions for unsteady problems	#1
dvdromnu New Member Join Date: Nov 2009 Posts: 4 Rep Power: 17	Hi, I am solving unsteady state convective diffusion scalar equation using finite volume method(explicitly). My problem is: For steady state problem the solution converges at some time step and we get results. That means steady state problem is independent of initial conditions. For unsteady state problem I am printing the values of scalar at a particular time step. So My question is, for unsteady problem what initial conditions I have to take..? means if I run unsteady problem for long time as in case of steady problem I am getting the same result of Steady state problem. So, how I have to solve unsteady problem. What initial conditions I should take. What is the difference between unsteady state and steady state solutions

November 7, 2009, 15:02		#2
alberto Senior Member Alberto Passalacqua Join Date: Mar 2009 Location: Ames, Iowa, United States Posts: 1,912 Rep Power: 36	If the problem admits a steady state solution, the unsteady solution procedure should lead to the same solution, if it is run long enough. __________________ Alberto Passalacqua GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM - An open-source implementation of quadrature-based moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using.

November 9, 2009, 01:49	solution profile	#5
dvdromnu New Member Join Date: Nov 2009 Posts: 4 Rep Power: 17	Actually I want to know whether there is any difference in solution procedure of unsteady state and steady state scalar convective diffusion equation. Also I want to know if we solve scalar equation with diffusion term only(without convection term) we will get one solution. Now, if we solve the same including convection term, we get another solution. So, adding convection term changes the solution profile( with only diffusion term) or it shifts the solution profile..?

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November 8, 2009, 01:45		#3
dvdromnu New Member Join Date: Nov 2009 Posts: 4 Rep Power: 17	Thank You................

November 9, 2009, 03:15		#6
dmoroian Senior Member Dragos Join Date: Mar 2009 Posts: 648 Rep Power: 20	Hello .... dvdromnu (actually it would be nice to have a name here), If the equations you're solving admit a solution, and if that solution is unique, then it doesn't matter what initial conditions you start with (as long as they are physical or representative for your equations). If you're interested in the steady state solution, then again it doesn't matter what approach you use (steady/unsteady). If the convection/diffusion equation has the same steady state solution as the diffusion only equation, then you will not see any "profile shift". The only difference would be in the choice of the solver, and the amount of computational time requested (shorter for steady state and longer for unsteady). I hope this was useful, Dragos