
[Sponsors] 
May 5, 2009, 11:39 
Some questions for steady and unstedy state

#1 
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Kostas
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Hi. I want some answers for these below:
1. If a problem flow is unsteady and i try to run it with steady state what happen? Will it converge sometime? 2. If i take convergence solution of a problem and after i run it (with this initial contition) with unsteady is there a case to appear unsteady effects? thanks 

May 5, 2009, 11:49 

#2 
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Danial
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Depends what you mean by unsteady effects.
One case can be the vortex shedding regime. You can have a steady state simulation of a cylinder in the flow. However, in case of vortex shedding, the results (steady state vs unst. st.) will not be the same (in fact deviate alot). However, you can use the st. st. as the initial condition for unsteady state. This is a good method if you are working on Fluid structure interactions, however I doubt it is helpful for CFD. The unsteady effects need to be initiated, such as vortex shedding in symmetric setup, which usually are initiated due to numerical perturbations in CFD (having a not too fine mesh, round off errors). hope it could answer your question a bit. 

May 5, 2009, 12:12 

#3 
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Kostas
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Consequently, when you run a steasy state, you will take a converged solution, regardless if the problem is, in fact, unsteady. Simply it wouldnt right solution with steady.. Right?
All these i write you because i want to learn for solution on my problem(hydrofoil with cavitation). Thus.. i run my model with a steady state and i take o converged solution... However, after, i wonder if my problem is steady. And thus i run this solution which i take with unsteady and i saw that all of time steps are same... Hereat.. the problem is steady! I am right??? 

May 5, 2009, 12:21 

#4 
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Danial
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It may be true or not.
It is possible that your solution of unst. st. is not correct, due to:  Not fine enough mesh, that can not resolve the local effects: Do a mesh study  Not small enough time steps: decrease your time by 10, 100.  Avoid a symmetric domain: if you have a symmetric hydrofoil, put it lower or higher than middle of the domain. this will initiate unst. st. as well Strouhal number + CFL + Reynolds number are your tools, so use them. 

May 5, 2009, 12:53 

#5 
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Kostas
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Yes.
My mesh must be ok ( i have much attention for this (y+ approximately 1, structured, dense). Now for time step i hve put 0.01 sec.. but i try and with 1sec or more.. But with these values dont i have instability of solution??? The end, this witch tell me i did not know ( with place of hydrofoil in domain). Why can be have more right solution?? 

May 5, 2009, 12:56 

#6 
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Kostas
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have you msn?? for talk more comfotably


May 5, 2009, 14:12 

#7 
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Danial
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not that way. What I meant was to divide your time scale.
1e2s, 1e3s,... If you have a cylinder, let's say it has a strouhal number of 0.2 (it is different for hydrofil), then: St = f.l/v 0.2 = f*1[m]/0.5[m/s] > then your frequency of vortex shedding should be 0.01 Hz, or each period should be 100s. this way your dt should be 1/20th of it to capture all the period (=5s here), given you are in vortex shedding regime (Re>50) 

May 5, 2009, 14:24 

#8 
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Kostas
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ok i understand.
You, if you want to solve a maybe unsteady problem what procedure will be you allowed? You will ru steady until converge or not? and after you will eun unsteady? 

May 5, 2009, 14:33 

#9 
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Danial
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If I want to run an unsteady model, first I will make sure that everything is fine by running a st. st.
When the results are fine, in flow problems, I will ramp the flow velocity slowly. It gives the dynamic problem an adaptation period. Let's say if your max velocity is 2m/s, I will start from 0 m/s and in 10s bring it to 2m/s. Usually it will not work if you start your simulation with very high velocities at t=0 (too steep numerically) 

May 5, 2009, 16:20 

#10 
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Kostas
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ok but if i have initial condition from steady state with a normal velocity? What? I will change again the velocity at v0=0?


May 5, 2009, 16:24 

#11 
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Danial
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no, either:
1 do st st, then do unst. st. and use st. st. as initial condition 2 start from unst. st. by using time ramping, v_inlet = 0>2m/s I prefer 2, choice is yours. 

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