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Time step ???

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Old   September 25, 2013, 14:11
Default Time step ???
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sud
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Hi ppl,
I have a very big question in setting the time step for my numerical simulation. We used to specify the time step in unsteady simulation. What does it mean actually ? .

I have a case where a body drops down. In this , i start the cfd simulation with time step being specified by stability criteria at certain height with u =100 m/s and t=0 s as initial condition. After several iteration , the solution converges . Now i take the total time step it took to converge . LEt it be t= 3s . Now by that time , this body has reached to some place in the domain with some loss in the velocity to about u=85 m/s . Now again with this thing as initial condition , can i proceed as above ?

Can i proceed with this procedure ? and try to calculate the temperature variation on that body starting at t=0 till some t=8 or 10 seconds .

Is this the correct way of solving ?

Thanks in advance:co nfused:
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Old   September 25, 2013, 17:13
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Filippo Maria Denaro
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Quote:
Originally Posted by archeoptyrx View Post
Hi ppl,
I have a very big question in setting the time step for my numerical simulation. We used to specify the time step in unsteady simulation. What does it mean actually ? .

I have a case where a body drops down. In this , i start the cfd simulation with time step being specified by stability criteria at certain height with u =100 m/s and t=0 s as initial condition. After several iteration , the solution converges . Now i take the total time step it took to converge . LEt it be t= 3s . Now by that time , this body has reached to some place in the domain with some loss in the velocity to about u=85 m/s . Now again with this thing as initial condition , can i proceed as above ?

Can i proceed with this procedure ? and try to calculate the temperature variation on that body starting at t=0 till some t=8 or 10 seconds .

Is this the correct way of solving ?

Thanks in advance:co nfused:
In your case I do not suggest to adapt the time step increasing it only because the velocity decreases. The time step is constrained by stability requirement that involves also the diffusion term of the equations. You could go in numerical instability
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Old   September 25, 2013, 17:42
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yes. The time step is constrained by stability requirements. But my question is just like a dynamic condition . For example take the spacehuttle , during reentry phenomena , the reentry dynamics is governed by dynamics equation which gives the trajectory , its velocity at each and every altitude and the time taken to reach the ground . When doing CFD simulation of the spacecraft throughout the trajectory, the initial condition changes at each and every seconds. Like that way , i have a body falling down with some initial velocity, u=100m/s at t=0 and lets assume that the body is at 100km alititude. Now to do the CFD , I take the free stream conditions at 100kms and use the unsteady governing equations (compressible or incompressible) and the solution converges after 1000 iterations. It took 1500 seconds to obtain this convergence (adding up all the time step each iteration. t=t+delta t).

Now at time is 1500 seconds let the decayed velocity of that body say u =85 m\s (obtained by solving the same reentry dynamics) . and altitude 60km . Can I carry out the same type of CFD simulation , taking u= 85m/s and t=0, and time step as per the stability and see the convergence.

The question is can i use the time taken to converge as a guide to pick up the velocity and altitude (from reentry dynamics) ????
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Old   September 26, 2013, 02:26
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Hi,
Try to use Galerkin method with upwind schemes (depends on Peclet number). In this case there is no problem with time step.
Marek
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