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June 19, 2008, 22:11 |
Steady State Vs Transient answers
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#1 |
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Hello, It may not sound a serious question, but I still need to clear my doubt.
If I run the same simulation in steady state solver and transient solver, and somehow I manage the simulation to converge in steady state solver. Will the result from both the simulations almost same? I could only manage the steady state simulation to converge by playing around the false time step. Thanks so much, Kushagra |
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June 20, 2008, 21:06 |
Re: Steady State Vs Transient answers
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#2 |
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No, we expect more accuracy from transient simulation because it is time marching solution. I am very new to CFD though, it will be interesting if more experienced CFD user can give us insight into this.
Regards |
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June 22, 2008, 19:51 |
Re: Steady State Vs Transient answers
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#3 |
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Hi,
The difference between a steady state simulation and marching a transient solution to steady state is that the SS simulation ignores many of the cross terms and higher order terms dealing with time. These terms all go to zero in steady state so they don't affect the steady state result. The transient simulation includes all these terms. Usually this means the steady state model has an easier convergence as there are less terms to model and some transient non-linearities are removed, but in a few models these non-linearities help convergence (but this is infrequent). Glenn Horrocks |
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July 3, 2008, 00:25 |
Re: Steady State Vs Transient answers
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#4 |
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Thanks Glenn for providing useful knowledge about convergence in both type of simulations. But what about quality of results from two type of simulations? Some time steady state simulations are difficult to converge but reducing the false time step by 1 or 2 order, they converge. Does it affect the quality of results?
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July 3, 2008, 02:55 |
Re: Steady State Vs Transient answers
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#5 |
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Hi,
A fully converged simulation, run to steady state by either steady state or transient approaches should be the same. The only exception is when "local timescale factor" is used in a steady state run as it can accelerate convergence nicely but as different timescales are used across the domain can cause accuracy problems. As long as a steady state run is run to final convergence with a physical timescale (including auto timescale) then it should be fine. The timescale is a steady state simulation is like under-relaxation from SIMPLE based solvers. Too high a URF and the simulation diverges, too low and convergence is slow, so you try to fiddle until you get the optimum in the middle somewhere. Glenn Horrocks |
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July 3, 2008, 23:50 |
Re: Steady State Vs Transient answers
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#6 |
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Thanks so much Glenn.
1) So reducing the 'false time step' is similar to reducing the under-relaxation factor. (Or is it just opposite?) 2) Suppose the residual for Volume Fractions are not getting low and fluctuating around mean values, what should the user try first? Reducing or increasing the 'false time step'? 3) The residence time (volume of domain / flow rate) is 13 second for my multiphase case. what might be a good time scale to start with steady state problem. Meanwhile, I found some of your, Robin's, Cyclone's replies on this forum about how the steady state solver works. They were really helpful. Thanks, Kushagra |
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July 6, 2008, 23:41 |
Re: Steady State Vs Transient answers
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#7 |
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Hi,
1) Be careful thinking of the physical timestep size as a false timestep. False timestepping is a different technique used in SIMPLE based solvers (I think). In CFX, a steady state solver is similar to the transient solver, just with some higher order transient stuff and cross terms removed, and a different residual calculation. But in basic idea, yes, tuning the URF of a SIMPLE run is similar to tuning the timestep size of a CFX run. There will be an optimum value somewhere between too slow and unstable. 2) http://www.cfd-online.com/Wiki/Ansys...gence_criteria 3) A good starting point is 13 seconds. Glenn Horrocks |
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July 11, 2016, 11:27 |
Transient and SS !
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#8 |
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Chaitanya
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Hello,
A basic conceptual doubt: I am performning a conjugate heat transfer problem for a drilling process. Lets assume, I want to see the temperature profile of the tool after 30s.. I have a tool fixed at intial temp. (lets say 600K). It loses heat by conduction and convection to the surrounding air. (radiation losses are neglected). I will perform a transient analysis setting appropriate time steps.. Will this result be the same as a steady state simulation freezed at 30s. ? (is this possible in CFX? Can i set it up the steady state simulation directly from 30s.) I would say, it wont be the same because of: 1. Hconv is proportional to delta T which is proportional to time. (different tempertures at different time steps). 2. the conduction equation is a parabolic PDE, which has time. so the heat loss due to conduction is different at different time steps. |
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July 11, 2016, 12:25 |
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#9 |
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Unfortunately, the answer is "depends".
If the transient calculation reaches steady state before 30 [s], say the transient contribution is already negligible at that time, the solution on the same mesh should be identical for a linear set of differential equations. For a non-linear system, there will only be identical if the system has one stationary solution. If the time to reach steady solution is larger than 30 [s], the transient solution will be very different than the steady solution. Summary: in general, they are very different unless you already know the "steady state time". Hope the above helps, |
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July 11, 2016, 12:40 |
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#10 |
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Chaitanya
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Thanks for the reply.
Lets say, I know the time when I achieve steady state.Lets say this time is 120 [s]. ( I assume I achieve steady state, when there is no change in heat trasnfer, momemtum trasnfer.... etc. with time for the same mesh.) At this instance, will my transient and steady state simulation yield the same results ? the fact is that I do not understand how CFX evaluates the steady state solution. If there is some Literature about this, share it. |
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July 11, 2016, 13:30 |
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#11 |
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At this point, you are better off reading the ANSYS CFX documentation on how either analysis is carried out, as well as reading a book on partial differential equations.
If your application is purely heat transfer, a heat transfer textbook usually cover heat conduction on both steady and transient modes. I would not worry much about the momentum equation since the concepts are unique. |
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July 11, 2016, 19:44 |
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#12 |
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Chaitanya
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Thanks for the references.
All I could find in the literature of CFX was the following informtion. photosphoto_1.JPG. photo_2.JPG. I have the solver guide and the CFX Pre guide. I do not know if there is some other literature that I can review. I searched various portals. Could you refer me some literature, about equations that are solved for steady state ? |
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August 16, 2016, 08:29 |
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#13 | |
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Quote:
Dear Glenn, I have tried to find out what are these "cross terms" but I didn't manage to find the good answer. Could you please explain what did you mean by this? Best regards, |
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August 16, 2016, 09:23 |
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#14 |
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Glenn Horrocks
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I vaguely recall that there are some neglected cross terms but I cannot find any reference to it or recall much about it. Bruno's comment here (http://www.cfd-online.com/Forums/cfx...transient.html) says there is no difference except in the way the coefficients are updated in the iterations. There are also differences in the way the residual is calculated.
So I am not sure about the cross terms. But the updating of the coefficients and residuals are certainly different. |
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August 16, 2016, 09:48 |
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#15 |
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Dear Glenn,
thank you for the reply, yes indeed, there is a difference in the way the RMS is computed, I have tried to find it in the documentation but I have not managed yet. As far as I know, for SS it is: RMS = | where the N is the number of elements. Do you know maybe whether it is treated in the Ansys documentation? Specifically for the transient case. I would like to find out more on this topic. Best regards, |
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August 16, 2016, 20:51 |
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#16 |
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Glenn Horrocks
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Yes, that is the standard definition of RMS. The residuals are defined in section 11.2.3 in the theory guide. The definition is not thorough, but enough to get the idea. It also defines the difference between SS and transient.
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September 21, 2016, 10:22 |
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#17 | |
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annn
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Quote:
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September 21, 2016, 19:55 |
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#18 |
Super Moderator
Glenn Horrocks
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I have changed my opinion on this since 2008. I use it much more often these days.
The main simulations where it makes a difference are things with complex coupling between equations, such as multiphase or shock waves, or where there is transient vortex shedding (which implies the simulation was not steady state in the first place). |
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September 22, 2016, 17:28 |
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#19 |
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annn
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Hi Glenn,
thanks for replying so fast! Just a follow up, do you think that if you modeled a steady state case (where you knew was suppose to be a steady state case) with a transient one there should be no problem as they should theoretically yield the same result if it was truly a steady state case? |
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September 22, 2016, 19:46 |
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#20 |
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Glenn Horrocks
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Yes, the transient simulation should converge over time to the steady state case.
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