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About the difference between steady and unsteady problems

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Old   June 24, 2000, 22:34
Default About the difference between steady and unsteady problems
  #1
Lisa
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Dear readers,

I'm solving 2-D driven cavity problems. I read several papers and found that some people used steady equations to solve this problem, but others used unsteady equations to solve it and even gave the animotion of their results. If physical problem is the same, why some get unique solution and some get time dependent solution? Can you tell me the reason.

Thank you.

Lisa
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Old   June 25, 2000, 10:11
Default Re: About the difference between steady and unsteady problems
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John C. Chien
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(1). The answer is very simple : "garbage_in_garbage_out". (2). The garbage here is just some objects, it is not the real garbage. (3). Having defined that, lets come back to the cfd. CFD deals with the numerical solutions of fluid dynamic equations and boundary conditions on computer. (4). In other words, the starting point is the governing equations and the corresponding boundary conditions. Then you try to use different method(S) to solve this set of equations through a computer program.( develop method, and write program) (5). Now the garbage is { equation(S), method(S), and program(S)}. (6). For example, { steady state equation, upwind method, program written with relaxation method} will produce a unique kind of solution (garbage). (7). On the other hand, {transient equation, central-difference method, program written with explicit method} will produce a sequence of time dependent solutions(garbages). (8). Now you know why there are different kinds of solution (garbage) available for the same geometrical similar problems. (9). For the lid driven 2-D problem, the problem can be either steady state, or transient. For example, the lid can be initially stationary, and then set to motion at certain rate until it reaches a preselected speed. To simulate such transient problem, the equations used must be transient. This is quite obvious. (10). Even if the problem is steady state, one can argue or speculate or assume (I hope that you are free to do so in the first place)that his program will produce a steady state solution eventually (long time solution will converge to the steady state solution). That is another kind of idea (garbage). (11). There is another kind of idea that without spending any money in cfd development, one can get a copy of off-the-shelf commercial code, and generate a non-uniform mesh (in a square domain for the cavity problem in this case) in a few minutes, then a solution will be there with clicking of some options in the GUI windows. This is another kind of idea (garbage). This is especially attractive to the non-technical managers. This is the death trap. Not just garbage. I will get into the details of this kind later, if I have the time in the future. (12). Well, if the steady state solution exists (for the kind of conditions specified), and if you are after the steady state solution, then you should use the steady state equations so that the solution will be steady state. (if such solution can be obtained ) (13). But, this does not prevent others (for example engineers in the same company) to propose that by using the existing unsteady transient code, or off-the-shelf transient codes, the same steady state solution can be obtained , thus save time and money for the company (the managers love this idea )(it is based on the garbage assumption, thus the idea is also garbage). (14). When the transient approach failed to produce a steady state solution, two possibilities will happen, one is to present the animation to show off the capability of the code and the effort involved, the other is to give the manager a copy of the transient solution at time t, which is likely to cost the manager's job later on when the product can't meet the customer's efficiency requirement. But that is a different story. But a real one though. (15). In the battle field of cfd war, you must keep calm and still be able to ask very simple and basic question like your question. I think, we are fighting a cfd war in a garbage dump, mountain high. (16). By the way, do we really have to solve this lid driven 2-D cavity flow? Yes, I think it is very important for our survival. Especially, if someone is spending your tax money on the assumption that there are indications of the presence of H2O on Mars. (that is higher order garbage idea) (17). At least, there are direct applications of the 2-D cavity flow in industries. (18). And if you have spent many years to develop a method to solve the transient flow equation, are you going to give up the opportunity to present your (garbage) idea? that by running the code long enough there will be a steady state solution? (instead of garbage?) (19). As a survival skill, don't ask this question, if your boss has a transient code. or if there is a large project to study transient behavior in such problems like multi-stage rotor-stator flow interactions. In that case, you should say his code will be very valuable in solving the fluid dynamic problems on Mars! GOOD GRIEF! So, don't put the garbage in your head in the first place.
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Old   June 25, 2000, 12:06
Default Re: About the difference between steady and unsteady problems
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Lisa
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Dear John C. Chien,

Thank you for your response.

I still have a question about this problem, If I calculate the case with Reynolds number as high as 20000 or higher, then this physical phenomena is unsteady, so I can't use steady equation to solve this problem, only unsteady equation can be used. is it right?

Thank you very much.

Lisa
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Old   June 25, 2000, 13:11
Default Re: About the difference between steady and unsteady problems
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John C. Chien
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(1). Well, a very good question again. (2). The lid driven 2-D or 3-D cavity flow has been used as a standard test case to test the performance of any new solution algorithm to solve Navier-Stokes equations. (3). It is perfectly all right to run a case at Re>=20000, using the steady-state formulation. It can be used to check the performance of various algorithms for this range of Reynolds numbers. And you should be able to obtian the steady state solution if the equation you are solving is steady state. (4). We can always come back and say that such solutions in this Reynolds number range do not exist in nature at all. It is a valid question. (5). But, remember that this lid driven 2-D cavity flow test problem is not a real one. It is a mathematical one for checking out the new algorithm. It is not a physical one. (6). For this reason, you can still use the steady state equations. The 2-D cavity flow solution is in no way real physical solution. (7). In order to validate the cavity flow solution by testing, it must be 3-D. And there will be limitations and complications in running the test. So, the test can only simulate the ideal cavity flow problem only. (8). Now if you are interested in the simulation of real physical problem, it is likely that the flow for this Reynolds number range (Re>=20000) will be turbulent. Thus you must solve the equations with a turbulence model. (9). The thing will become more complicated, that is whether the mean turbulent flow in the cavity is steady state, or un-steady. It is likely that steady state solution with a turbulence model will be adequate. But that same steady state solution can always be obtained by someone using the transient turbulent flow approach. (10). To the other extreme, one can try to obtain the solution by using the transient DNS approach. In this case, in principle, you don't have to worry about whether it is laminar of turbulent. (11). AT that point, you will be completely lost. I mean, "what is the original goal of running the case of 2-D lid driven cavity flow?" Are you developing the DNS technology? or Are you checking out the transient flow algorithm? or Are you developing the turbulence model? or Are you trying identify whether the flow is steady or unsteady in this Reynolds number range? or Are you checking out whether the algorithm can produce a steady state solution in this Reynolds number range? (12). In CFD, the requirement is the reliable and reproducible solutions consistent with the assumption and equations used. Whether the same condition and solution exist in nature, is a separate question. Even with the testing, it is hard to say whether flow at Re=20000 is going to be steady, unsteady, laminar or turbulent. After all, these are not relevant questions when one decide to run a 2-D lid driven cavity flow case. (13). So, the answer to your question is: you can either use the steady state equation or use the transient state equation. The choice is yours. (14). But one thing you have to remember is, numerically, there is no guarantee that the long time solution of the transient equation will converge to the steady state solution. So, if you are looking for the steady state solution, and the transient solution keeps changing, then you don't have a solution at all. You can't say that the solution is likely to be transient. (15). In other words, transient solutions can hide a lot of effects in it, for example, the initial guess of the flow field, the mesh quality, the accuracy of the algorithm used etc... (16). To check out the performance of a new algorithm, the steady state euqation is recommended. Otherwise, it is up to you.
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Old   June 25, 2000, 15:12
Default Re: About the difference between steady and unsteady problems
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Lisa
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Dear John C. Chien,

Thank you very much, you helped me a lot.

Lisa.
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Old   June 25, 2000, 16:20
Default Re: About the difference between steady and unsteady problems
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Lisa
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Dear John C.Chien,

I have ever used 3-D time dependent NS equations to simulate the flow field around delta wing. I got the following results, the solution is steady when angle of attack is less than critical angle( before vortex breakdown), Increasing the angle of attack ( exceeding critical angle of attack, vortex breakdown occurs), I can't get steady solution, the solution is time dependent (transient solution). Can I say my simulation reflects the basic real physical phenomena( suppose my methods are correct)? In mathematical aspects, Is the solution changed from steady to unsteady caused by the angles of attack related to the Hopf bifurcation or chaos?

Lisa
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Old   June 25, 2000, 19:03
Default Re: About the difference between steady and unsteady problems
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John C. Chien
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(1). In general, when there is no streamwise flow separation, the flow is stable. This is the case for the flow over a delta wing where the the flow separation is in the cross-stream plane vortex with the vector pointing in the streamwise direction, when angle of attack is small. (2). As soon as the streamwise flow separation occur, as in the flow after the vortex breakdown, flow oscillations similar to the wake flow of a cylinder will occur. (3). Flow over a body at high angle of attack will create very complex separation patterns on the surface and in the flow field. So, I must say that, in general it is similar to the real flow behavior. (4). But then, it depends on the Reynolds number, the wing leading edge shape, and Mach number,etc... I mean whether it is initially laminar, or turbulent. (5). In additions, the votex pair can be either symmetric, or asymetric, depending upon the angle of attack. (6). What I am trying to say is that just because the solution has turned into unsteady motion, we can not say that the solution is real. There are several other factors need to be considered also. But in any case, it is the first step in the right direction.
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Old   June 25, 2000, 21:34
Default Re: About the difference between steady and unsteady problems
  #8
Lisa
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Dear John C. Chien,

In your last response, in item 5, you wrote that the vortex pair can be either symmetric, or asymetric, depending upon the angle of attack. Yes, in my calculation, when the angle of attack reached a certain value, the vortex pair changed from symmetric to asymetric. Although I got the result similar to the experiment, but I don't understand that the same equation and boundary condition, and the geometry of delta wing is symmetric, why is the vortex pair asymetric?

Thank you

Lisa
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Old   June 25, 2000, 22:24
Default Re: About the difference between steady and unsteady problems
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John C. Chien
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(1). It is fairly common in the missile aerodynamics, that is flow over a slender body at high angle of attack.
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Old   June 26, 2000, 02:46
Default Re: About the difference between steady and unsteady problems
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Duane Baker
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Hi Lisa,

it is actallly very common with strongly forced flows (in fact any nonliner problem with strong forcing eg. hi Re fluids, slender beams and shells under high loading, etc) with symmetric geometry, b.c's and source terms can lead to asymmetric solutions. One thing to note is that for these ideal conditions the asymmetric solutions often come in pairs or higher multiples. A numerical algorithm will often tend to one of these multiple solutions because of assymetries in the algorithm ie. iterating the i,j,k from 1,1,1 etc. The real physical solution will often have a preference due to asymmetries in the physical problem (often very tiny) which have been neglected in the model. For example, a small scratch in the wall of a diffuser will cause the flow to separate off of one wall or the other. Often the preference for a physical solution is not the same as the one from the code. This takes quite a bit of investigation.
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Old   June 26, 2000, 17:31
Default Re: About the difference between steady and unsteady problems
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Lisa
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Thank you very much !

Lisa
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Old   July 5, 2000, 14:37
Default Re: About the difference between steady and unstea
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Adrin Gharakhani
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If you solve an unsteady problem with a transient code and time-average the results (given you have let the simulation proceed through statistically sufficient number of cycles) you will get an answer similar to the so-called steady state solution. What is crucial to recognize here is the term time-averaged, which unfortunately many researchers corrupt by terming it "steady state", either due to ignorance of physics or lax attitude toward using the correct terminology.

Adrin Gharakhani
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