| John C. Chien
|
March 31, 1999 12:32 |
Re: unsteady calcs in FLUENT
(1). If you are solving the transient equations, the solution obtained will be a function of time ( and a function of space also). (2). If the problem you are trying to solve takes an hour physical time (not the computing time) to finish ( I mean to reach the steady state), then you can divide it into 60 steps and each state will represent one of the transient solution . But if you divide it into two steps, you are not going to see a lot of the transient evolution of the solution. (3). Since most currently available methods are based on finite-difference, finite-volume, or finite-element appraoch, the solution is only approximate unless you reduce the cell size, or time-step size to a very small number ( approach zero ). Based on this principle, the time step has to be infinitesimal and the number of time steps will reach infinity. Now you have a real problem. (4). So, the practical way to determine the time step is to work backward. First, determine the physical time required for your problem to reach the steady state ( you got to know something about your problem first). Then, determine the computer time required for the code to finish one time step ( you also need to know something about the black box you are using). Then, determine how much physical time you are going to spend on this problem. That will determine the total number of time steps within this period of time. Say, the code takes one minuite to finish one time step of calculation on your problem, and you can spend only one hour on your problem,then the total number of time steps will be sixty. Now, you have determine the total number of time steps,that is 60.(5). Back to your problem, if the physical time for your problem to reach the steady state is one second, then the physical time step will be (1 second/ 60 ). Now you have determined the physical time step. So, as you can see that it is up to you to determine the total number of time step and the physical time step size required as the input to the code. No one else can answer the question for you. (6). There will be time when the code you are using has additional limitations, such as time step size stability requirement (CFL or thing like that), inner iteration loop inside the time step (for example, the pressure-based method), etc. In that case, you will have to follow the user's guide to check your time step and the number of inner iteration steps to see whether they satisfy the requirement or not. (7). If you are solving an oscillating flow problem, you can determine the total physical time to cover several cycles ( instead of just one period).(8). Your question is very similar to the question of " what kind of watch should I wear?" For me, I am wearing a watch with only one hour hand and one minuite hand only. In some cases, you may want to have additional second hand in the watch, or even a stop watch type hand. So, the decision is yours. (9). If your main interest is the steady-state solution, then you don't have to worry about the definition of the time step at all.
|
