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How to determine time step size and Max. iterations per time step. 

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June 21, 2013, 14:16 

#21 
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Tanjina Afrin
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And yes, when I select noniterative time , these are the default solution method.


June 23, 2013, 05:13 

#22 
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Yash Ganatra
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I will try to help. You can post more details


June 23, 2013, 06:44 

#23  
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Sijal Ahmed Memon (turboenginner@gmail.com)
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Because for each situation you will have different flow physics/solver requirment and which dictates the time step size and no of time steps. First few general rules: 1. It is better to lower the time step instead of increasing the no of iterations 2. Lowering the time step will enhance convergence. 3. Use better initial conditions (get from steady state solution) 4. If you experience difficulty in getting convergence for desired time step, use lower time step for initial transients. once you get good convergence gradually increase time step to required value. 5. Adaptive time stepping will give you faster and automatic transient simulation to desired time step. 6. Use 2nd order implicit scheme. Now few situations: 1. LES, explicit and implicit schemes. For LES, the condition for time step selection is based on cournt number and CLF < 0.2. For explicit scheme, used when you have advantage of faster convergence for flows where flow delta T is of same order as dictated by CFL otherwise use implicit scheme. For explicit scheme delta T is determined by CFL < 1 by stability constraint. For implicit scheme delta T is determined by the flow feature you are interested in. 2. General time step calculation for arbitrary flows Delta T = (1/3) * (L/V) or lower value 3. Turbomachinery : (1/10) * ( number of blades/rotational velocity) I typically use this time step calculation method for sliding mesh problems: total time to pass one blade pitch (if you have equal no of stator and rotor blades) = pitch in meters / velocity in m/s Now If you want to resolve it in 20 steps, time step would be total time / 20. For better resolution if you want to use 100 steps per wake passing, time step would be total time/100. 4. Vortex shedding: Time step can be found from strouhal number in indirect manner. From strouhal number (fix the density, viscosity and velocity and known value of diameter of chord length) find out the frequency of vortex shedding. Inverse of it is the total time of vortex shedding period. Lets we have f = 0.2 therefore total time is 5s. Now if use atleast 25 time steps to represent vortex shedding sequence we will have the 5/25 = 0.2 sec for each time step. That is delta T = 0.2 seconds Judging the convergence, time step dependence You can judge convergence by plotting surface monitor of variable of interest at particular boundary, line or point. If solution starts to have periodic behaviour, then you have reached convergence. What should come first : Mesh independence or time step independence study? My vote goes for mesh independence for sample flow conditions. then go for time step dependence study. Post precessing of unsteady flows: Post processing can be done in Fluent, CFDpost and tecplot. For nicer pics, I prefer tecplot. For post processing of some variable, you need to turnon data sampling for time statistics . If you want some additional variable you should make the custom field function and enable it in sampling option panel for "custom field function" For post processing you should run simulation for long enough time and when you start to observe the periodic solution, then turn on data sampling. And take data sample for alteast 58 cycles. 

June 23, 2013, 22:20 

#25 
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Tanjina Afrin
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@ Far,
Yup, when I select Non iterative time advancement, these are the default settings except "scheme" where default setting is PISO, I made it fractional step manually. "4. If you experience difficulty in getting convergence for desired time step, use lower time step for initial transients. once you get good convergence gradually increase time step to required value." how can I do it ? @Yashganatra, I am trying to model a flow through a circular pipe from a rectangular reservoir. I used pressure inlet.At first reservoir is full of water. I want that water level will be down with time since no water will enter into reservoir and air will replace the place of water. I used workbench for meshing. It's triangular mesh. Element size is 6e3m. attached image is showing the result after 1400 run ( when I used water volume fraction 0 in inlet). result was almost same for 2000 run, 500 run... 

June 24, 2013, 04:56 

#26  
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Sijal Ahmed Memon (turboenginner@gmail.com)
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June 24, 2013, 07:01 

#27 
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Yash Ganatra
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Hi Tanjina
Due to the pressure difference water will flow down, the outlet is open to atmosphere? From the image, time elapsed is 1.4e3, it's very less. Maybe you will need multiphase or a UDF needs to be written as was said above. 

June 24, 2013, 09:45 

#28 
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Tanjina Afrin
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@Yash ganatra,
Since I am using workbench and mesh shape is triangular , I don't understand what will be the mesh size. From body sizing, I use element size 6e3 m and I used it as del x. using courant number , I found del t=0.001, but when I use this time step, it says divergence in Xmomentum , then I reduce relaxation factor for momentum from 1 to 0.1, 0.7 , 0.3 etc. after some run, it showed some other problem ... so I use very small time step. for this small time step, it takes lots of time to complete 20003000 run. Should I use 500010000 run with this time step ? @ Far, I am using VOF multiphase model. I don't know how to write UDF . It will be great if you suggest any site from where I can learn writing UDF. And simultaneously I am searching for UDF learning site. " for example if your required time step is 0.001 and you have convergence problems in beginning of solution, then just reduce time step to 0.0001 or 0.00001 and once solution is stabilized, change time step to 0.001 gradually. " I still don't understand this part . After starting the "run calculation" how can I change the time step without stopping the run ? And it is possible with 0.00001 time step I run 1000 run, then next 1000 run I will continue with large time step ? @Yashganatra and Far, I am a very new user of Fluent, may be I am bothering you so much..... I am sorry for that. 

June 24, 2013, 13:30 

#29 
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Tanjina Afrin
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Should I use "open channel flow" option here ? But it's not a open channel flow, is it ?


June 25, 2013, 01:30 

#31 
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Yash Ganatra
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Tanjina,
Even I am a beginner so not a problem. For writing UDF, download FLUENT UDF Manual. It is comprehensive. Also it has a section on basics of C Programming Yash 

June 25, 2013, 17:16 

#32 
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Tanjina Afrin
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@ Far and Yashganatra,
Thanks you guys a lot. My model is working now ( in 2D) without any UDF. Trying to run in 3D. Attached is the image of residual of my model . Shouldn't it be converged ? Please enlighten me. 

June 26, 2013, 01:29 

#34 
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Yash Ganatra
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That's great. What changes did you make?


July 3, 2013, 09:54 

#35 
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Tanjina Afrin
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@ yashganatra, I didn't check the "implicit body force" and "operating density condition" , after checking this two point, model gave the expected results. Thank you all for your kind help.


July 15, 2013, 17:11 
perforated pipe time step.

#36 
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Tanjina Afrin
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Hi all,
I tried to model a perforated pipe in 2D, thats why I make some opening and flow was okay at the beginning, when I make time step 0.0001, though the residual was like attached image. then I increase time step time gradually, when I make it 0.01, after some iteration, it just hanged up. What can I do? Before hang up, the image was like the attached image 2. Any help will be really appreciated. 

January 7, 2014, 01:33 
water hammer simulation

#37 
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mohamed abdelrahman
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Hi all
am on my last step on my master degree and iwoud like to simulate water hammer in pipe in cfx line coud any one give me more datails about that itray before but ididn't get good reslt .... plz 

August 28, 2014, 05:49 

#38 
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Thomas Green
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Dear All,
I have the same problem of DIVERGENCE DETECTED IN AMG SOLVER: ymomentum. I am simulating a constant flow profile but I would like to analyze it in transient mode. My mass flow inlet is of 0.000001 kg/s, I thought that using an adaptive time step that change from 0.001 for the first second to something bigger for the rest of the simulation (for instance 10 seconds) because the fluid flow will be developed. Solution Method (in order): PISO (Skewness Correction & Neighbor Correction = 1), Least Squares Cell Based, Second Order, Second Order Upwind, First Order Implicit. Run Calculation: Adaptive Time Stepping Method (deltat); Time Step Size (s)=0.001 (because I have to select the initial time step); Number of Time Step=1000, Max Iterations/Time Step=200. My code for the adaptive time stepping is: #include "udf.h" DEFINE_DELTAT(deltat,d){ real time_step; real tx=CURRENT_TIME; if ( tx<=1 ) { time_step=0.001; } else{ time_step=10; return time_step; } } I don`t get very well what is wrong in what I am doing, anybody has suggestions? Thank you so much. Thomas 

February 11, 2015, 07:44 

#39 
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enass
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If i have a case file how can i know the smallest cell size?


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